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Quantum gravity in an accelerated frame of reference It is said that we can't study quantum gravity because gravity is a weak force. But gravity and acceleration are the same. Why can't we study quantum gravity in a strongly accelerated frame of reference?
| This is a classic case of equivocation. The word "gravity" is used in the question with two related-but-different meanings:
*
*When we say "gravity is weak," we mean that a lot of stuff (like matter) is needed in order to give spacetime a noticeable amount of curvature.
*When we say "gravity and acceleration are the same," we mean that the experience of standing on the surface of the earth is locally indistinguishable from the experience of accelerating (being pushed from your feet toward your head) in flat spacetime. Conversely, the feeling of weightlessness you have when drifting in deep space is the same as the feeling of freely falling toward the surface of the earth with no resistance at all.
Distinguishing between those two meanings of the word "gravity" is the key to answering the question. We have always known (since before I was born, at least) how to formulate quantum physics in a prescribed spacetime background, with or without curvature, and with or without acceleration. That's how Hawking radiation was originally derived: using a prescribed spacetime background. When people say that quantum gravity is difficult, that's not what they mean. The difficult part is accounting for the fact that quantum stuff causes spacetime curvature. Things like Hawking radiation and the Unruh effect are about what acceleration and/or spacetime curvature does to quantum stuff. The hard part of quantum gravity is understanding the converse: what quantum stuff does to spacetime curvature.
| {
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Why the mechanism of everything in the universe has a pattern? why everything in the universe has a pattern which can be identified and understood to determine outcomes, properties, effects of almost everything. I am saying that couldn't the universe be like patternless, non-deterministic and chaotic. For example why the gravitational force between any two objects has a pattern which always obeys universal law of gravitation and can be predetermined. Couldn't be the gravitational force between any two given objects would have no pattern and would be completely random and non-deterministic. Is this property of universe in which everything has a pattern is a complete matter of chance or it is a property of even something fundamental.
| As a matter of fact, we do have a quantifying parameter for chaos, which is Entropy.
In terms of entropy our universe is chaotic, but for better or for worse, this chaos seems to increase monotonically with the passage of time, which is a predictable feature, unlike what you would want. Maybe in future, its possible that someone would come up and say that everything that is predictable about this nature stems out of this peculiar feature. As of now, this is an open problem commonly called as the Arrow of time.
Coming to you question as to why Predictability -
If the universe weren't predictable, as in absolutely chaotic, that would mean that entropy has reached a maximal state, there cannot be any more chaos!! Now, unchanging parameter of a system (including entropy) is the characteristic of equilibrium state. As a matter of fact, if everything where to be in equilibrium, NO LIVING SPECIES CAN EXIST. We feed ourselves each day in a strive to maintain a NON-equilibrium state, that's what keep up alive - CONSTANT INPUT OF ENERGY.
Hence, the very existence of living species can be attributed to the fact that we are living in a non-equilibrium world.
So essentially my claim is that, although as you have imagined, we cannot eliminate the possibility of a universe with absolutely no pattern (and such a universe may exist somewhere), but for us its sufficient to know that the universe we live in is predictable and we know that through BIOLOGY, for one thing.
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Will two bodies initially connected to and revolving around each other, start spinning when disconnected? Two extended bodies are connected with a string and revolve around each other (that is, around the center of mass of this system). No gravity, no external forces.
The string is cut, and they start to depart from each other. Will they spin around their own centers of mass?
For a (representative) particular case when one of the bodies is a dumbbell (two point masses connected by a light rigid rod),
the answer is trivial - each point mass will start its tangential movement with a different velocity, so the pair will be revolving.
On the other hand, for the following configuration,
there seems to be no rotation.
Is there a general approach?
| The bodies are spinning before the string is cut, once per revolution. Since neither the string nor the cutting of the string apply torque to the bodies, there will be no change of rotational speed for the bodies, so they will keep spinning.
To think of this another way: at the moment the string is cut, the part of the body connected to the string is moving slower than the part of the body directly opposite due to the circular motion. Cutting the string does not change the speed of any part of the body, so the difference in velocity in different parts of the body will result in an overall rotation.
In your second illustration, the two balls have different velocities, as shown with the arrows below:
These different velocities would result in an overall rotation in the two-ball system after the string is cut.
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Do we hear sound at pressure or displacement antinodes? I have read from Young & Geller (2007), College Physics 8th Edition, Pearson Education Inc. (pg 385) that we hear sound at pressure anti-nodes rather than displacement anti-nodes as microphones sense pressure variations.
When we have 2 speakers facing each other to form a standing wave between them, it forms a standing wave with open ends at both speakers. This means that the ends closest to the speaker are displacement antinodes and thus pressure nodes. However, going by the above fact that we detect loudest sounds at pressures antinodes, this would mean that when we place a microphone closest to the speakers, it should detect very soft sounds. Yet, many experiments online showing standing wave formation between 2 speakers seem to show that the loudest sound will occur closest to the speakers.
So where exactly do we hear/ detect sound?
Thank you!
| I was visiting a science center which had an acoustic standing wave set up in a long tube through which people could crawl. The standing wave pattern was drawn on the side of the tube and it looked to me that they had made a mistake be showing nodes at the two open ends however, when I did crawl through the tube, I found the ends were quiet and there regularly spaced loud and quiet places along the tube. So I can confirm that my ears at least detect pressure nodes and antinodes rather than displacement ones.
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Density operator in different bases This may sound like a silly question, and it probably arises from a weak understanding of the concept.
What I was wondering is: given a density operator $\varrho=\sum_{i,j}p_{ij}|i\rangle\langle j|$, can we find separate bases for which $\varrho$ is diagonal, but presenting different eigenvalues?
Mainly I was considering the example of when $\dot{\varrho}=0$. In this case $\varrho$ is diagonal in the Energy eigenbases: this grants us a nice representation of $\varrho$ as we can clearly observe its eigenvalues for each energy level. My question would then be: is possible to find a second bases(not necesseraly an observable eigenbasis) for which $\varrho$ becomes diagonal but presenting different entries?
Intuitively I would think that this should obviously possible as long as some commutation relation is respected ($=0$), but the whole idea is becoming quite confusing to me.
| If I understand your question correctly, then the answer is no. The eigenvalues of an operator - which are simply the diagonal entries when that operator is diagonalized - are uniquely defined up to trivial reordering. Furthermore, the vectors belonging to distinct eigenspaces are orthogonal because $\rho$ is self-adjoint and therefore symmetric.
so if I am given a diagonal density operator, and then someone claims that ϱ˙=0, then I know for a fact that the ϱ I was given is being represented in the energy basis, right?
No, not quite. If two energy eigenspaces appear in the density operator with the same probability, then you might use a basis in which $\varrho$ is diagonal but $H$ is not.
For example, consider the 2-level Hamiltonian and density operator in some basis $\{|1\rangle,|2\rangle\}$. $$H=\pmatrix{E&0\\0&-E} \qquad \varrho = \pmatrix{0.5 & 0 \\ 0 & 0.5}$$
Obviously $[\varrho,H]=0 \implies \dot\varrho = 0$. However, if we change to a new basis
$$|1\rangle = \frac{|a\rangle+|b\rangle}{\sqrt{2}} \qquad |2\rangle = \frac{|a\rangle-|b\rangle}{\sqrt{2}}$$
then the Hamiltonian and density operator become
$$H = \pmatrix{0 & E \\ E & 0}\qquad \varrho = \pmatrix{0.5 & 0\\ 0 & 0.5}$$
As a result, even though $\dot \varrho = 0$, we have a basis $\{|a\rangle,|b\rangle\}$ which is not an energy eigenbasis, but in which $\varrho$ is diagonal.
This is a nice reminder that $[\varrho,H]=0$ implies that there exists a basis in which $\varrho$ and $H$ are simultaneously diagonalized; however, this does not imply that every basis which diagonalizes one also diagonalizes the other.
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Light: reflection and transmission When white light hits a thin "transparent material", is there a general relationship between the spectra of the reflected and transmitted components?
For example, do red filters (that make things look red when viewed through them) also look red when light is shined upon them? Or is this not necessarily true?
| "White" light is a bit ambiguous. Let's just call it a mixed spectrum of wavelengths in the visible spectrum for now.
Now, for a given material, assuming it is semitransparent, there is a spectrally varying reflection coefficient. The light that isn't reflected has the complement to that curve (highly reflected light means very little is not reflected, etc.). Then apply the spectral absorptance curve to see what light makes it to the second surface. Once again, apply the spectral reflectance curve, essentially repeating the spectral loss curve of the first surface.
So, in the case of zero absorptance, the reflectance curve and the transmittance curves will be "opposite" to each other. I'm leaving out the possible etalon behavior due to the internally reflected light going back to the first surface, partially reflecting/transmitting there, and so on for innumerable round-trip reflections.
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How do I decompose an $SO(10)$ antisymmetric tensor in $SO(4) \times SU(3) \times U(1)$? My guess is that If I denote the $SO(4)$ indices $\mu, \nu = 1,...4$ and the $SU(3)$ indices by $I,J=1,2,3$, I think $N^{mn}$ should decompose as $N^{\mu \nu}, N^{IJ}, N^{I}_J, N_{IJ}$ plus other terms with mixed indices $I,\mu$, which I don't know how to determine.
I would appreciate if someone could give an explanation of an honest way to decompose $SO(10)$ in $SO(4) \times SU(3) \times U(1)$.
| Hints:
*
*Firstly,
$$SO(10) ~\supseteq~ SO(4)\times SO(6),\tag{1}$$
so we get the branching rules
$$ {\bf 10}~\stackrel{(1)}{\cong}~({\bf 4},{\bf 1}) \oplus ({\bf 1},{\bf 6}),\tag{2}$$
and
$${\bf 10}\wedge{\bf 10}~\stackrel{(2)}{\cong}~ ({\bf 4}\wedge{\bf 4},{\bf 1})\oplus ({\bf 4},{\bf 6})\oplus ({\bf 1},{\bf 6}\wedge{\bf 6}).\tag{3}$$
Here the space of 2-forms
$$ {\bf 4}\wedge{\bf 4}~\cong~{\bf 3}_+\oplus{\bf 3}_-\tag{4}$$
splits into selfdual/antiselfdual parts under 4D Hodge duality.
*Secondly,
$$SO(6)~\supseteq~ U(3)~\cong~[SU(3)\times U(1)]/\mathbb{Z}_3,\tag{5}$$
cf. e.g. this Phys.SE post. Up until now the vector spaces under point 1 could in principle be real. In particular the representations are real. Now we complexify the vector spaces. We get the branching rules
$$ {\bf 6}~\stackrel{(5)}{\cong}~{\bf 3}_1 \oplus \bar{\bf 3}_{-1},\tag{6}$$
and
$$\begin{align} {\bf 6}\wedge{\bf 6}
~\stackrel{(6)}{\cong}~&({\bf 3}\wedge{\bf 3})_2
\oplus (\bar{\bf 3}\wedge\bar{\bf 3})_{-2}
\oplus ({\bf 3}\otimes\bar{\bf 3})_0 \cr
~\cong~&\bar{\bf 3}_2 \oplus {\bf 3}_{-2}
\oplus ({\bf 8}_0 \oplus {\bf 1}_0)
,\end{align}\tag{7}$$
cf. e.g. this Phys.SE post.
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What is the intuition behind the spacetime interval? In an article that I am currently reading (under the Lorentz Invariants sub-heading), it explains that, just as the distance between two points on a Cartesian plane are obviously invariant of the coordinate system, the “spacetime distance” is also invariant. While in Cartesian coordinates $$(x_1-x_2)^2+(y_1-y_2)^2 = (x_1'-x_2')^2+ (y_1'-y_2')^2,$$ the space time analog is $$c^2(t_1-t_2)^2-(x_1-x_2)^2 = c^2(t_1'-t_2')^2- (x_1'-x_2')^2 = s^2$$ where $s^2$ is the spacetime interval.
I am having difficulty in understanding this notion of a spacetime interval and the intuition/derivation for why it can be written in this way and is invariant under Lorentz transformation.
I am aware that similar questions have been asked on this platform but none of them have fully cleared things up for me so far. Any help in providing an intuition or understanding would be appreciated.
| The intuition is not too difficult to build. For convenience, I will write $\Delta x$ instead of $x_1-x_2$.
First, we know from the first postulate that the speed of light is invariant. If we write $$\Delta x^2 + \Delta y^2 + \Delta z^2 = c^2 \Delta t^2$$ this is the equation of a sphere of radius $c \ \Delta t^2$. In other words, this is a flash of light moving at $c$ in all directions. We can rewrite this easily as $$0 = -c^2 \Delta t^2+\Delta x^2 + \Delta y^2 + \Delta z^2 = s^2$$ where in this case $s$ is fixed to 0.
Now, written this way $s^2=0$ is clearly invariant by the first postulate. It is invariant under spacetime translations and spacetime rotations (which includes boosts). And with $\Delta t=0$ we know from our experience with Cartesian coordinates that $s^2=a$ is invariant under spatial translations and rotations. So it is not a big intuitive leap to think that $s^2=a$ is invariant under spacetime translations and rotations.
So, let’s take a step back and think about what we know and can guess. We know by the Pythagorean theorem that $\Delta x^2+\Delta y^2+\Delta z^2$ gives us an invariant in space. We want to figure out what function of time $f(\Delta t)$ we need to add so that $s^2=f(\Delta t) +\Delta x^2+\Delta y^2+\Delta z^2$ is an invariant in spacetime. We know by the second postulate that $f(\Delta t)=-c^2 \Delta t^2$ is the right form for the special case of $s^2= -c^2 \Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2 =0$, so it is intuitive to think that is probably the right form for all $s^2$, not just zero.
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Induced emf in a rotating disc A question in my book was given as
A conducting disc of radius R rotates about its axis with an angular velocity $\omega$. Then the potential difference between centre of mass of disc and its edge is (no magnetic field is present)
The answer stated that:
*
*There will be a potential difference present
*The value of which was obtained by using $eE = m\omega^2r$ (not sure why this holds true) for an electron $e$ in the disc and then obtaining the electric field and finding the potential difference by integration -$E.dr$
My question: While deriving the motional emf in a rotating disc in magnetic field, we consider only the flux change (or say the $(v\times B).dl$) responsible for the induced emf given as $\frac{1}{2}B\omega r^2$. Why do we not consider the emf obtained due to the above phenomenon aswell since it happens even in the absence of magnetic field?
Extra but related question: Why does $eE = m\omega^2r$ hold true for an electron in a conducting disc? What stops it from moving radially?
| Both these questions have been asked (and answered) within the last 3 or 4 weeks on this site. The gist of the answer to your main question is that the 'centrifugal emf' is very much smaller than the 'magnetic emf'. You should show this for yourself by putting figures into your formulae. [Note that a magnetic field of 1 T would be a very strong field.]
To answer your extra question: the free electrons do move outwards at first, as there is no force to provide them with a centripetal acceleration. But as they move outward they create a charge separation: negative to the outside of the disc; positive to the inside. These charges create an electric field, $E$, that does provide the required centripetal acceleration. Hence the equation that you've quoted: $=^2$.
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Intensity of reflexion of circular polarization Circularly polarized light, can be caracterized as :
$$E(r,t) = E_0 \cos(kz-\omega t) \hat{x} + E_0 \sin(kz-\omega t) \hat{y}$$
In my course, the Intensity of EM radiation is defined as $I = |U|^2$.
And we defined polarized monochromatic waves as follows (i already put the phase for (right) circular): $$U(r,t) = (E_x \hat{x} + E_y \hat{y} e^{i \pi/2})e^{i(kz - \omega t)} $$
Now, i can get the first equation by taking the real part of the second one.
But now, let it reflect of some glass and let's assume the s and p polarizations coincide with the x-y axes. How do i get to $I = I_0(R_s + R_p)/2$. Where the $R_s$ and $R_p$ are the reflectivities.
The Instensity as defined in the course doesn't seem well defined to me. Do i take the sum of the instensity along each axis (But then i miss a factor of $1/2$) ?
| Yes, you take the sum of the intensity along each axis. But you do not miss the factor of $1/2$ by doing that. The reason is that the total input intensity to the reflection is $I_0=U^2=|E_x|^2 + |E_y|^2$ and $|E_x|=|E_y|$; Along each axis the input intensity is ${I_0}/{2}$. With reflection reflectivity $R_s$ and $R_p$, your output intensity is $\frac{I_0}{2}R_s+\frac{I_0}{2}R_p=I_0(R_s+R_p)/2.$
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Newtonian vs Lagrangian symmetry Suppose we have a ball of mass $m$ in the Earth's gravitational field ($g=const.$). Equation of motion reads as:
$$
ma = -mg
$$
From here we can conclude that we have translational symmetry of the form $x(t) \to x(t) + const$ (we are working in only 1D). However, we cannot see this symmetry from the Lagrangian:
$$
L = \frac{mv^2}{2} - mgx
$$
because the linear term "breaks" this symmetry. Moreover, we also do not have the corresponding conserved quantity (as far as I can see).
Does this mean that we can have symmetries in the Newtonian sense (transformations that map solutions to other solutions) that are not present in the Lagrangian?
| You can do an integration by parts on the last term (and discard the resulting boundary term) to yield an action with equivalent EOMs:$$
S'= \int \left( \frac{1}{2} m \dot{x}^2 + m g t \dot{x} \right) \, dt
$$
In this context, the symmetry $x \to x + C$ is obvious at the level of the Lagrangian. Moreover, the Euler-Lagrange equations become
$$
\frac{d}{dt} \left( m \dot{x} + m g t \right) = \frac{\partial \mathcal{L}'}{\partial \dot{x}} = 0
$$
and thus the quantity $m \dot{x} + m g t$ is a constant of the motion. Specifically, it's the initial momentum of the particle.
(This seems "cheap", somehow, and I'm not 100% sure whether it's a legitimate move. Comments are welcome.)
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Are Newtonian forces real forces or just a useful construct? I am new to physics but not new to science/scientific thinking.
Since I was young I have never really understood how to interpret the Newtonian forces. In some cases they seem very real. E.g. the static friction force can be nicely explained as gluing of atoms, or something like that.
But if you consider two objects of different weights on the same table, then the table applies 2 different normal forces to the 2 objects. This really seems to be an abstract construct. We see that the objects are not moving so to be self-consistent with this massive framework of forces we are building we are claiming that the table is exerting a force towards the object. Yeah sure! What kind of atomic interpretation could this have?
Of course, it does work at the end, so I could totally accept the "construct" interpretation, since it's a useful model anyway. But what transpires from reading articles etc. is that Newtonian forces are "more" than that. They are actually "real".
So, what is the right way to conceptualize the Newtonian forces?
I hope I expressed my confusion clearly.
|
We see that the objects are not moving so to be self-consistent with
this massive framework of forces we are building we are claiming that
the table is exerting a force towards the object.. Yeah sure! What
kind of atomic interpretation could this have?
Everything we see around us is made of atoms (mostly organised in molecules) In turn, atoms consist of a tiny, positively charged nucleus surrounded by a much, much larger negatively charged electron cloud.
All electrons are negatively charged and negatively charged objects repell each other by the Coulombic force.
This is what physically underpins the force and reaction force when a massive object sits motionlessly on a table (e.g.): the electron clouds of the atoms making up the table and the electron clouds of the atoms making up the massive object repell each other.
As an interesting aside this also means that the table and object never really touch and that the object kind of hovers a very small distance above the table's surface.
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Is there a physical reason behind electromagnetic energy and momentum being derived from the Lorentz force equation? The effect of the EM field upon a charge $q$ is given by the relativistic Lorentz force equation:
$$\frac{dP}{d\tau}= qF^{\alpha\beta}U_{\beta}$$
The expression on the RHS is then substituted via Maxwell's equations to define EM field conservation laws and their corresponding definitions of field energy and momentum within them. Is there a physical reason for defining the energy and momentum of the electromagnetic field this way?
| The relativistic Lorentz force equation ensures that the four force $\frac{dP^\alpha}{d\tau}$ is a contravariant four vector. In particular, the quantity
$$
\frac{dP^\alpha}{d\tau}\frac{dP_\alpha}{d\tau}=\left(\frac{dP^0}{d\tau}\right)^2-\left(\frac{dP^1}{d\tau}\right)^2-\left(\frac{dP^2}{d\tau}\right)^2-\left(\frac{dP^3}{d\tau}\right)^2
$$
is the same in every Lorentz frame. Physically this means that an electron moving in a magnetic field experiences the same force when we observe it in the frame where it is at rest. In that frame the force comes from an electric field. This was in fact the starting point for Einstein to develop SR.
The discussion here is closely related.
Note that the electron experiences only the spacial components of the force which in isolation are not Lorentz invariant. However this is approximately true at low speeds.
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Why is the angle between the radial velocities at two instants the same as the angle between the tangential velocities at those same instants?
While it is clear that the angle between $v_r$ and $v_r + \Delta v_r$ is $\Delta \theta$, I cannot see a clear geometric reason as to why the angle between $v_t$ and $v_t + \Delta v_t$ must also be $\Delta \theta$.
The book by Kleppner and Kolenkow uses this to argue that $\Delta v_t \approx v_r \Delta \theta$ and in the limit in which $\Delta t \rightarrow 0$ , $\frac{d v_t}{dt} = v_t \dot{\theta} = r\dot{\theta}^2 $.
| The statement is obviously true for motion in a circle. For any short arc (swept out in a short time) a center of curvature can be found so that the arc can be considered part of a circle.
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What will be the angular velocity of a line about a line that is at some angle with the plane of motion? In this book it is written that angular velocity of a rigid body is time derivate of "angular displacement" of any line in the plane of motion of the body. The angular position of the line is measured from any conveniently fixed reference axis in the plane of motion.
For example in the figure shown below, the authors have chosen the horizontal line to be the reference axis. The angular positions of lines $1$ and $2$ are $\theta_{1}$ and $\theta_{2}$ respectively.
I don't understand how this can be the definition used to find the angular velocity of the lines (1 and 2), which move in the plane, from a reference line that make some angle to the plane of motion. What will be angular positions for lines 1 and 2, from this angled reference axis? What is angular displacement?
For concrete example suppose in the figure below I take $OM$ as reference axis and I want to find the angular velocity of the line $OP$ that is rotating anticlockwise maintaining the same height $PM$ and $NP$ (although they don't look to be the same in my very bad drawing). In starting $OM$ and $OP$ makes angle $\alpha$ at some time and then after infinitesimal seconds later the line moves and now makes angle $\beta$ with the line $OM$.
So should the angular displacement be $\beta -\alpha$? This seems very wired; how is this difference related to the rotation?
I have read many books on this topic but they all give the same definition which involves the reference axis to be in the plane of motion. It seems that we can't take any line as a reference axis.
| If a solid object is rotating about fixed axle, all points (and lines) in that object will circle the axle in the same period of time. (They have the same angular velocity.)
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Why do icebergs flip over? Why do icebergs flip over? Are certain shapes of icebergs more "stable" than others, in that it's harder to flip them over? If so, why?
For example, it somehow makes intuitive sense, that a thick iceberg with a certain height (or depth, because 90% of it is below water) would be harder to flip over than a thin one with the same height/depth. But why is this the case?
| This is a nice Newtonian physics question. A very simplified analogue to this is to consider the stable positions of a 2-dimensional table shape iceberg with length $L$, width $W$ ($W < L$).
Typically the material density ratio of $\rho_{\text{ice}} / \rho_{\text{water}} = 0.9$. In a force equilibrium, this leads to the fact that 90% of the iceberg is immersed in water. As with the inverse pendulum, there are stable and unstable force equilibriums.
Algebraically, it is simpler to consider the energy equation for a table iceberg with the short side up versus the long side up. You'll find that the short side up is energetically better, and leads to a stable minimum.
Most of the energy difference is dissipated in creating water waves during the flip.
| {
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Why isn't time reversal a Galilean transformation? I'm a mathematician learning physics from scratch, starting from Newtonian mechanics. As far as I understand, Galilean transformations are defined as transformations of space-time that transform from one inertial frame to another. In turn, an inertial frame is a frame of reference in which Newton's first law holds: a body not acted upon by any force will move in linear motion. From those two definitions, it seems like Galilean transformations should be all transformations of space-time that preserve linear motion.
However, Galilean transformations are then described as compositions of:
*
*Translations of space and time: $(\mathbf{x}, t)\mapsto(\mathbf{x}+\mathbf{x}_0, t+t_0)$.
*Orthogonal transformations of space: $(\mathbf{x}, t)\mapsto(R\mathbf{x}, t)$ where $R:\mathbb{R}^3\rightarrow\mathbb{R}^3$ is an orthogonal transformation. Note that it includes both rotations and reflections of space.
*Rectilinear motion: $(\mathbf{x}, t)\mapsto(\mathbf{x}+t\mathbf{v}_0, t)$.
All of them definitely preserve linear motion, but they are not the only ones. We also have:
*
*Linear transformation of space: $(\mathbf{x}, t)\mapsto(T\mathbf{x}, t)$ where $T:\mathbb{R}^3\rightarrow\mathbb{R}^3$ is an invertible linear transformation.
*Time stretching: $(\mathbf{x}, t)\mapsto(\mathbf{x}, at)$, $a>0$.
*Time reversal: $(\mathbf{x}, t)\mapsto(\mathbf{x}, -t)$.
I see why the first two are problematic: they do not preserve measurements of distances and time intervals (so the definition of galilean transformations should really mention that too...). But what's wrong with time inversion? It doesn't seem to be any more "problematic" than reflections of space. Both preserve quantitative measurements but invert the orientation.
| It certainly seems inconsistent to include spatial reflections but at the same time to exclude time reversal.
One reason for excluding both spatial reflections and time reversal from the definition (the convention followed in Wikipedia, for example) is that this means the topological group of Galilean transformations has a single connected component.
| {
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Can an electron near thar the positive terminal of a battery make it all the way to the positive terminal? Following Drude's model of conductivity, can an electron near thar the positive terminal of a battery make it all the way to the positive terminal? The electron is in the wire but near the positive terminal. At least, is it probable? Assuming that the circuit size is commonly small. The reason I asked this is that I do now know how to interpret the average velocity of an electron in this model. Because I know it bounces off impurities until reaching a stationary average velocity. Can I use this average velocity to estimate a time to go from one side to the other using the formula of uniform rectilinear motion?
| Maybe, if the circuit is small enough, and runs for long enough.
The velocity of an electron through a wire is approximately 1mm/second. As a result, if you have a circuit 1cm long, and you let it run for 10 seconds, then an electron that starts off at the negative end of the circuit would reach the positive end of the circuit.
Note that this velocity is different to the speed of the electrical current, which is approximately the speed of light; the reason why it's so much faster is because each moving electron creates an electric and magnetic field around itself that in turn pushes the electron in front of it forwards, which pushes the electron in front of it, and so on and so forth until the entire column of electrons is moving almost instantaneously.
| {
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When a person pulls or pushes a cart, why is it advantageous for their body be tilted forward? This is not a homework question. I attempted to draw a free body diagram for a person pulling or pushing a cart.
Based on Newton's third law, the following forces act on the body of the person:
*
*forward reaction force done by the ground because of friction between the person and the ground.
*downward force (the person's weight) done by the earth.
*backward reaction force done by the cart.
I am wondering why the body of the person must be tilted forward. I have not seen any relationship between this posture with the magnitude of the forces acting on the person.
Could you tell me why the person's body must be tilted forward? How does this posture provide mechanical advantages?
| It's similar to why when you pull something with a rope along the ground it can never be vertical. It needs to have a horizontal component to apply a horizontal force.
This is not quite as true with rigid structures like legs, but it similar in that it reduces bending stresses on the leg bones and torques on you knees and hips. It effectively reduces the lever arm formed by our legs between our hips and feet which increases the force that can be applied. A wheel (a round lever) is an example of where this does snot happen: a straight lever that pushes parallel to the ground while being perpendicular to it.
If you calculate the internal forces on the leg bones and the force/torque at the joints, you will find the differences. What you're doing right now is similar to calculating the force at the edge of a large and small wheel and being unable to find the difference. The difference is the torque at the shaft.
| {
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When a car accelerates relative to earth, why can't we say earth accelerates relative to car? When a car moves away from a standstill, why do we say that the car has accelerated? Isn't it equally correct to say that the earth has accelerated in the reference frame of the car? What breaks the symmetry here? Do the forces applied to the car have special significance in determining which frame is inertial and which one is not?
Please explain in simple terms.
| Assuming the acceleration of the earth is immeasurably small due to its vastly larger mass, I think your confusion actually stems from this:
Isn't it equally plausible to say that the Earth has accelerated from the point of view of the reference frame of the car?
It depends what you mean by the reference frame of the car. In the inertial reference frame in which the car was stationary, the car is now moving, whereas previously it wasn't. It is the car that has accelerated.
In the reference frame in which the car is now stationary, and the earth is moving backwards, the car was previously moving backwards, so it is still the car that has accelerated.
The observer who sees the earth apparently accelerate, does so because she has changed reference frames, or to put it another way she is in (or has been in) an accelerating reference frame. This is where the asymmetry comes from, because this observer is accelerating with the car and not viewing the system from any inertial reference frame.
| {
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Find the values of $A$, $B$, and $C$ such that the action is a minimum
A particle is subjected to the potential $V (x) = −F x$, where $F$ is a constant. The particle travels from $x = 0$ to $x = a$ in a time interval $t_0 $. Assume the motion of the particle can be expressed in the form $x(t) = A + B t + C t^2$. Find the values of $A, B$ and $C$ such that the action is a minimum.
I was thinking it can solved using Lagrangian rather than Hamilton. There's no frictional force.
$$L=\frac{1}{2}m\dot{x}^2+Fx$$
$$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0\implies m\ddot{x}=F\implies \ddot{x}=\frac{F}{m}$$
Differentiate $x(t)$ twice. $$2C=\frac{F}{m}\implies C=\frac{F}{2m}$$
For finding B I was thinking to integrate $\ddot{x}$ once.
$$\dot{x}=\int \ddot{x} \mathrm dt =\ddot{x}t$$
initial position is 0 so, not writing constant.
$$\dot{x}=\frac{F}{m}$$
Differentiate $x(t)$ once.
$$B+2Ct=\frac{F}{m}$$
$$\implies B=\frac{F}{m}-\frac{2Ft}{2m}=-\frac{Ft}{2m}$$
Again, going to integrate $\ddot{x}$ twice.
$$x=\iint \ddot{x} dt dt=\frac{\ddot{x}t^2}{2}$$
initial velocity and initial position is 0.
$$x=\frac{Ft^2}{2m}$$
$$A+Bt+Ct^2=\frac{Ft^2}{2m}$$
$$A=\frac{Ft^2+Ft-F}{2m}$$
According to my, I think that C is the minimum (I think B is cause, B is negative; negative is less than positive). And, A is maximum.
A person were saying that "It asked you to minimise the action; it told you the particle moved from $0$ to $a$ in time $t_0$; it gave you the equation of the trajectory."
In my work where should I put the interval?
| I suspect that this is not the approach that the problem author wants you to take. Were I confronted with this problem, I would do the following:
*
*Determine what the conditions $x(0) = 0$ and $x(t_0) =a$ tell me about the parameters $A$, $B$, and $C$. I should be able to eliminate two of them in favor of the third using these two equations.
*Plug the resulting form for $x(t)$ into the action and integrate it from $0$ to $t_0$.
*Differentiate the resulting quantity with respect to the remaining parameter to find the value of the parameter which minimizes the action.
One should, of course, find that $C = F/2m$ after all this.
| {
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Relation between potential energy and conservative force Does potential energy only happen when the work done is by a conservative force? Or does work done by non-conservative forces also create potential energy?
| For a conservative force the following property is true
$$\oint_C \vec F \cdot \vec {dl}=0$$
Using stokes theorom:
$\oint_C \vec V \cdot \vec{dl}=\iint_S (\vec \nabla \times \vec V)\cdot \hat ndA$ where $\hat n$ is a normal vector which gives
$$\vec \nabla \times \vec F=0$$
as a matter of fact if $\vec F$ can be written as a gradient its curl will always be zero(curl of a gradient field is always 0) .If a fields curl is 0 we can find a potential $\phi$ such that $\vec V=\vec \nabla \phi$.
For a conservative force such field is $-V$.
Which means $$\vec F=- \vec \nabla V $$
and V can be recognized as potential energy of the force. Which means that any conservative force has a potential energy and vice versa.
| {
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Why and when can the Earth be considered an inertial reference frame? The question has been asked (e.g., here and here), but I would like to get a more definitive and mathematically formal answer.
The Earth rotates around its axis, around the Sun, and participates in larger scale motions as a part of the Solar system. Yet, we often can get by treating it as an inertial reference frame (e.g., when constructing furniture, cars and buildings). In some cases we do need to take account for the effect of its rotation - e.g., in weather prediction one takes account for the Coriolis force, but we still consider the Solar system as the inertial reference frame.
We do that because:
*
*accelerations that we deal with (notably $g$) are much greater than the accelerations due to the other motions that it is involved in?
*we can neglect the non-inertial forces because all the objects around experience the same accelerations due to these forces?
*something else?
I am looking for a mathematically motivated answer. I also suggest delineating between what is specific to Earth (accidental) and what would apply to all (or most) planets/stellar bodies.
Update
I took the liberty to summarize the opinions expressed so far in my own answer. Yet, there remains non-inertial effects not covered by free fall and the Earth's rotation - those related to the Earth's finite size and responsible for the tidal forces (more specific question is here). Thus, this question still needs a canonical answer.
|
The components of the earth rotation at the latitude $\lambda$ are:
$$\mathbf \Omega_E=\begin{bmatrix}
0\\
\cos(\lambda)\, \Omega\\
\sin(\lambda)\,\Omega\\
\end{bmatrix}$$
from here we can obtain the pseudo forces due to the earth rotation
$$ \mathbf F_s=m\,\big(2\,(\mathbf\Omega_E\,\times \mathbf{\dot{R}})+\mathbf\Omega_E\times\,(\mathbf\Omega_E\times \mathbf R)\big)
$$
where $$\mathbf R=\begin{bmatrix}
x \\
y \\
z\\
\end{bmatrix}\quad,\text{is the postion vector}$$
$$\frac{\mathbf F_s}{m}= \left[ \begin {array}{c} \left( \left( \cos \left( \lambda
\right) \right) ^{2}x+ \left( \sin \left( \lambda \right) \right)
^{2}x \right) {\Omega}^{2}+ \left( -2\, \cos \left( \lambda \right) {
\dot{z}}+2\, \sin \left( \lambda \right) {\dot{y}} \right) \Omega
\\ \left( - \sin \left( \lambda \right) \cos
\left( \lambda \right) z+ \left( \sin \left( \lambda \right)
\right) ^{2}y \right) {\Omega}^{2}-2\, \sin \left( \lambda \right)
\Omega\,{\dot{x}}\\ \left( \left( \cos \left(
\lambda \right) \right) ^{2}z- \cos \left( \lambda \right) \sin
\left( \lambda \right) y \right) {\Omega}^{2}+2\, \cos \left( \lambda
\right) \Omega\,{\dot{x}}\end {array} \right]
$$
The EOM's in rotating system free falling
$$\begin{bmatrix}
\ddot{x} \\
\ddot{y} \\
\ddot{z} \\
\end{bmatrix}=-\frac{\mathbf F_s}{m}-\begin{bmatrix}
0 \\
0 \\
g \\
\end{bmatrix}\tag 1\\
g=\frac{G\,M_E}{(R_E+z)^2}=\frac{M_E}{\big(R_E\,(1+\frac{z}{R_E})\big)^2 }\approx \frac{G\,M_E}{R_E^2}$$
where $~R_E~$ is the earth radius , $~M_E~$ earth mass and G the gravitation constant
with $~\Omega=7.2710^{-5}~$[1/s]$~~,\lambda=\frac{40}{180}\,\pi~$
$$\frac{\mathbf F_s}{m}=\left[ \begin {array}{c} 0.000000005285290000\,x- 0.0001113828620\,{
\dot z}+ 0.00009346131845\,{\dot y}\\ -
0.000000002602497284\,z+ 0.000000002183754512\,y- 0.00009346131845\,{
\dot x}\\ 0.000000003101535488\,z-
0.000000002602497284\,y+ 0.0001113828620\,{\dot x}\end {array}
\right]
$$
thus $\frac{\mathbf F_s}{m}\approx \mathbf 0~$ and equation (1) is the inertial case.
| {
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Why exactly is the resistance of a conductor inversely proportional to the area of its cross-section? Before I explain my query, I would like to clarify that I am a ninth-grader who got this question while studying the formula $R \propto \frac{1}{A}$ where $A$ is the area of cross-section.
I have often asked this question to my teachers and they always give me the classic "corridor and field example". They told me that if 5 people walk in a corridor, they will find it harder to get across than if they were to be walking through a field- the same goes for electrons passing through a conductor. My counter-argument would be that if the width of the conductor increases, so will the number of positive ions (my textbook says that positive ions in conductors hinder the flow of current) and hence, more the resistance.
I would really appreciate it if the answer could be explained to me in simple terms as I'm not well versed with the more complex formulae involved in this concept. If not, do let me know of the concepts I should read about (preferably the specific books) to understand the solution better.
| Sorry for my poor english. My native language is french.
We can also ask the question about the conductance: why is it proportional to the surface.
The reason is that we are working with a model in which the current density is distributed uniformly over the cross section of the conductor. It is as if we had a lot of identical conductors in parallel with each other. So, for the same voltage, if we multiply the area by N, we also multiply the current by N. This would not be the case if the current distribution was not uniform. For example, if we take into account the skin effect.
Another interesting example is that of the Hagen-Poiseuille's law in hydraulics: the flow rate is proportional to the pressure difference but the hydraulic conductance is proportional to the square of the area: it is not proportional to the area ! This is because the velocity profile is parabolic (and not uniform). Doubling the pipe area is not the same as taking two pipes.
| {
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Could dark matter be explained by the fact of empty space expanding much faster than space inside galaxies? We know the expansion of the universe is accelerating and mass slows time.
So pockets of empty space experienced more time since the big bang.
That mean the current expansion rate of empty space is much faster and had more time to expand.
Could dark matter be the outside pressure of empty space on the galaxy?
Does the current model takes into account this phenomenon?
| No.
We observe the effects of dark matter within galaxies and across galaxy clusters, as well as across the entire universe. So it is by no means just about the rotation curves of galaxies, and your idea can not explain the data.
Yes, current models of structure formation properly take this effect into account - by ignoring it, since indeed it is negligible anyway.
| {
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Hartree type diagrams in Altland Simons We are told in equation 5.25:
$$F^{(2), 1} = - \frac{T^3}{L^6} \sum_{p_1, \, p_2, \, q} \; G_{p_1} \; G_{p_1 + q} \; G_{p_2} \; G_{p_2 + q} \; V(q)^2$$
$$F^{(2), 2} = \frac{1}{2} \frac{T^3}{L^6} \sum_{p, \, q_1, \, q_2} G_p \; G_{p - q_1} \; G_{p - q_1 - q_2} \;\; G_{p - q_2} \; V(q_1) \; V(q_2)$$
The text tells us:
... only configurations where all momentum arguments carried by the Green function are close to the Fermi surface contributes significantly to the sums (5.25). Considering the first sum, we see that, for small $|q|$ and $|p_i| \simeq p_F$, this condition is met, i.e. there are two unbound summations over momentum shells around the Fermi surface. However, with the second sum, the situation is less favourable. For fixed $|p_1| \simeq p_F$ [I think $p_1$ should be $p$], fine-tuning of both $q_1$ and $q_2$ is necessary to bring all momenta close to $p_F$, i.e. effectively one momentum summation is frozen out.
The text argues that the second equation in equation 5.25 has more fine-tuning necessary than required for the first equation in equation 5.25 in order to argue that at high densities, we can neglect the second equation.
What confused me is that it seems that the same amount of fine-tuning is required for both the first and second equations. In particular, for the first equation we need small $|q|$, $|p_1| \simeq p_F$, and $|p_2| \simeq p_F$. Don't we just need for the second equation small $|q_1|$, small $|q_2|$, and $|p| \simeq p_F$?
|
In particular, for the first equation we need small $|q|$, $|p_1|≃p_F$, and $|p_2|≃p_F$. Don't we just need for the second equation small $|q_1|$, small $|q_2|$, and $|p|≃p_F$?
This is correct. But notice, while small $|q|$ requires fine-tuning in all 3 directions in k-space, the condition $|p_1|\simeq p_f$ requires fine-tuning only in one direction - perpendicular to the Fermi surface. In other words, if the relevant scale of $|q|$ beyond which $V(q)$ can be neglected is $\Delta q$, and the thickness of the relevant states around $p_F$ is $\Delta p\approx \Delta q$, then the first sum, $F^{(2),1}$, goes over the volume $\Delta V_1=(4 \pi p_F^2\Delta p)^2 \Delta q^3$, while the second sum goes over the volume $\Delta V_2=(4 \pi p_F^2\Delta p) \Delta q^6$. The ratio of the two
$$
\frac{\Delta V_1}{\Delta V_2}=\frac{4 \pi p_F^2\Delta p}{\Delta q^3}\gg1.
$$
Hope, this helps.
| {
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Distances in General Relativity, near massive bodies I've been studying the basics of General Relativity, and my question is: does it make sense to say that near a black hole (or any massive body), distances increase, the way on a topographical map the contours get closer and closer near a steep hill? So that an object escaping from near the event horizon is actually having to travel thousands of times more than the 'real' distance to reach a destination far away. This seems to jive with the idea that an infalling object (from an outsiders perspective) seems to travel an infinite distance, fading away but never crossing the horizon, and also with the Shapiro time delay phenomenon. But I want to ask in case this is a false analogy. And if it is true, how would that "stretched distance" (i.e. all the local distances on the trajectory added together) relate to the distance seen by an outsider? How could the outsider ever say "I am 100 million miles from a black hole," if any test particle would actually travel many times greater distance to traverse it?
And how could a distant observer measure distance to a black hole anyway?
How would a distant observer measure a valid distance to any object, if not by sending and receiving a light ray?
| You can form $\int ds$ along a radial line at some given $t$ and thus create a notion of ruler distance. The ruler distance from the horizon is finite. You can see this also by looking at Flamm's paraboloid: distances along the paraboloid are finite.
However the time required for a particle to move outwards from the horizon is infinite.
| {
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Can one make already separate atoms/electrons entangled? Is it possible to make two separate atoms or electrons entangled on a quantum level? Not photons, those I know can be entangled already
| As soon as two atoms or electrons interact with each other, they become entangled.
Say you shoot two atoms against each other in a collision experiment. At $t=0$ the atoms have a definite momentum $\hbar k$, $|\Psi \rangle = |k\rangle |-k \rangle$. After the collision, they will be in a superposition of different momenta, but the interaction is such that the total momentum must be conserved. Hence the resulting state will look roughly like this
$$ |\Psi \rangle \approx c_1 |k_1\rangle |-k_1 \rangle + c_2|k_2\rangle |-k_2 \rangle +...$$
which is an entangled state. If you want to understand why the result looks like this you should study quantum scattering theory.
| {
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Normal force shift for ball kept in cavity
Adapted from JEE advanced paper-1 of 2020
If you see the left side of the Planck which the ball touches, it seems so that as we vary $\theta$, the contribution of the force from that point of contact drops to zero. I have marked the point of interest in paint:
As we reduce $\theta$ , we see both the edges touching the sphere contribute to supporting it's weight. But, as we increase $\theta$ it drops off. It is intuitive to understand for me, but I can't give a precise reason why as to this happens. Hence, the question:
For what exact reason does the normal force shift to the right edge as we vary $\theta$?
| If we assume that the two sides can exert a force only normal to ball, and that they are the only forces on the ball other than gravity, then this becomes just a linear vector solution.
The two forces must sum to zero out gravity and simultaneously have net zero horizontal force. This can only happen if the "downhill" side exerts more force than the "uphill" side.
Another way to say the same thing is that as gravity pulls the ball down, the forces from the downhill side most directly oppose this force, while the uphill side is not able to act in that direction. So the pull downward results in strong reaction forces on the low side and weak ones on the high side. In the limit where the bottom force acts vertically through the center of mass, there will be no force pulling the ball into the plank and no force from the uphill edge on the ball.
| {
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What is the meaning of complex distances in relativity? In flat Minkowski space with the metric signature (-,+,+,+) the spacetime interval $ds^2$ is defined as
$ds^2 = -c^2 dt^2 + x^2 + y^2 + z^2$
This interval (or better the square of the distance) can become negative (for timelike intervals) so the spacetime distance $ds$ can be complex.
In quantum mechanics a complex quantity is something which cannot be measured. Only real quantities can be measured. Does that mean that the spacetime distance becomes unmeasurable if it becomes complex? Or how is the interpretation in relativity for complex spacetime distances?
What confuses me is that one can also choose a different metric $(+,-,-,-)$. In that case the spacetime interval is
$ds^2 = c^2 dt^2 - x^2 - y^2 - z^2$
Then it is not timelike distances that become complex, but instead spacelike distances will be complex. So it seems to depend ob the convention of the metric signature when the space-time distance will be complex. So do complex spacetime distances (or negative spacetime intervals) have any physical meaning at all?
| The scalar $ds^2:=dx^\mu dx_\mu$ is real, but its sign changes when you switch metric convention. Whether $\sqrt{ds^2}$ is real or imaginary is therefore not purely a physical question.
On the other hand, if $ds^2\ne0$ then $dx^\mu$ satisfies either $dt=0$ or $dx^i=0$ in a suitable reference frame, and which is applicable is very much physically relevant. If we can impose $dx^i=0$, the path is causally traversable; if we can impose $dt=0$, the path's endpoints can be regarded as simultaneous events, making them outside each other's light cone. (But you probably know all that.)
If you're interested in the real-complex distinction, it can help to work in proper time $d\tau^2=dt^2-d\vec{x}^2/c^2$, which settles the convention question. Traversable paths have $d\tau\in\Bbb R$; between events we can take to be simultaneous, $d\tau\in i\Bbb R$. But in the latter case, we don't normally consider it convenient to work with proper time.
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Confusion on Normal force and its resolution on inclined plane I had the following doubt regarding normal forces and its resolution on inclined plane. (Also I am quite new to SE, so I have no idea how do we upload diagrams to better represent my arguement).
Consider two cases, in both cases there is a inclined plane with some angle of inclination ($\theta$), on which there is a block of mass m that is kept. In Case I, we fix our coordinate axes such that the x-axis lies on the base of the inclined plane, and the y-axis perpendicular to x-axis, and in the next case the x-axis is placed along the inclined plane and the y-axis is perpendicular to the x-axis.
The problem I was facing was that if we consider the normal force in case I and say that the normal force is $mg cos(\theta)$ then this is usually accompanied by the reasoning that there is no acceleration in the direction of normal force, therefore net force in that direction is 0, and hence acceleration is 0 as well.
But in this coordinate plane,saying that acceleration normal to the inclined plane is 0, is just like saying that its rectangular components is also 0, because if we have a vector whose magnitude is 0, then its rectangular components must also have a 0 magnitude. So in one sense we will be saying that the horizontal component of the block's acceleration is also 0, which is not true.
But in the second case, where we have a inclined coordinate system, we can easily make such a assumption that normal force is mgcos($\theta$) because in the inclined coordinate system, due to the normal force being perpendicular to the x-axis (which we placed on the inclined plane), it doesn't have any rectangular x-component and thus we can easily conclude the normal force to be mgcos($\theta$).
I don't understand how different coordinate system can predict different normal forces? Where exactly am I going wrong?
|
So in one sense we will be saying that the horizontal component of the block's acceleration is also 0
This would be true if you were to replace 'horizontal component' by 'contribution to the horizontal component'.
However there is an acceleration down the plane, and that has a horizontal component.
| {
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What is light cone? Explain to mathematicians who understand the Lorentz group but not light cone Mathematically the Lorentz group is precisely the $O(1,3)$ is the 4-vector rotation preserving the inner product of 4-vector under this metric
$$
\eta_{\mu \nu}=(+1,-1,-1,-1).
$$
There are four distinct sectors of this $O(1,3)$.
Say the 4-vector is $A$ and $B$, then for the Lorentz transformation $R_{n \nu} \in O(1,3)$ on the 4-vector, we have the following invariant inner product of 4-vectors under the Lorentz transformation:
$$
A^\mu \eta_{\mu \nu} B^{\nu} ={A^\mu}' \eta_{\mu \nu} {B^\nu}'=A^\mu R^{T}_{\mu m} \eta_{m n} R_{n \nu} B^{\nu}
$$
Lorentz group $O(1,3)$ explains the symmetry group of the spacetime at any fixed point.
Built in on this data of Lorentz group $O(1,3)$, how do we explain what is light cone to mathematicians?
Namely what is light cone? mathematically? How to explain to mathematicians who understand $O(1,3)$, but not light cone?
What exactly is this cone of light cone defined mathematically?
| The light cone is definied to be the set of 4-vectors
$(ct,x,y,z)$ satisfying
$$c^2t^2 - x^2 - y^2 - z^2 = 0.$$
Or written in covariant notation
$$\eta_{\mu\nu} x^\mu x^\nu = 0.$$
(image from Einstein for Everyone - Spacetime)
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Where the reaction force from Newton's third law is acting on this body? In the diagram below I have a fire extinguisher sitting upright on a skateboard. Gas is being expelled out of the fire extinguisher and causing the skateboard to move forward (which is to the right in this picture). This would work in a vacuum as well I am told. My understanding is that in Newton's third law when a body Y is pushing on a body Z then body Z is pushing back on body Y in the opposite direction but equal in magnitude.
You often see big arrows like the white ones in my diagram that illustrate Newton's third law. In my example though, where is the force B actually occurring and what causes it? I guess I'm still confused. There is gas which pushes the gas in front of it on the way out of the fire extinguisher and in return the gas in front pushes back on the gas behind it which causes a net force like force B? That's the best explanation I can think of. So again I guess my question is: where is the force B actually occurring and what causes it?
| The action force is the substance being pushed out of the extinguisher to the left of the diagram, labelled "force A". The reaction force is then the same substance pushing back on the extinguisher in the opposite direction labelled "force B", and since it is connected to a skateboard,
this force will cause the skateboard to move to the right of the diagram.
| {
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Do I have to apply more force than gravity to lift my leg above the ground? How much force do I do apply when I lift my leg above the ground? The same amount as gravity does on my leg (mg)? Or MORE than it (greater than mg)? If the displacement from the ground to my lifted leg is h meters, then what's the work done? mgh or more than mgh? Basically I wanna know if I have to apply more force or do more work than gravity to lift my leg up.
| Firstly laying out my physics/math assumptions
*
*no friction
*only moving the leg vertically (as if you are lifting a bag)
*constant gravity, $g$
*force on leg, $F$, only varies with time
*for $t<0$ the ground provides the force to stop movement
therefore we can limit movement to 1 dimension, $x$
and from newtons second equation we get $a(t)=-mg+F(t)$
Now we want to have the leg to start from rest and finish at rest, if it doesn't finish at rest then you haven't finish lifting it.
If $F(t)$ is a constant (either greater than $mg$ or less than $mg$) then we won't get this behaviour, it will either not start moving ($F=mg$) or not stop ($F\neq mg$).
But if we have a slightly more complicated expression for the force, for example
where $t_0$ is chosen so that the leg stops at the right height (which i'm going to call $L$).
$$F(t)=\begin{cases}
2mg & 0\leq t<\frac{t_0}{2}\\
0& \frac{t_0}{2}\leq t\leq t_0 \\
mg & t_0\leq t
\end{cases}$$
ie lifting the leg with twice the force of gravity for the first half then letting gravity slow it to a stop for the last half, then providing a constant force to match gravity so that it stays there. (a point of note is that the average force
is $mg$)
hence the acceleration is
$$a(t)=\begin{cases}
+mg & 0\leq t<\frac{t_0}{2}\\
-mg & \frac{t_0}{2}\leq t\leq t_0 \\
0 & t_0\leq t
\end{cases}$$
from this we can find the velocity, by either integrating or recognising that these are the same conditions as newtons equations of motion so would follow the same behaviour
$$v(t)=\begin{cases}
+mgt & 0\leq t<\frac{t_0}{2}\\
-mg(t- \frac{t_0}{2})& \frac{t_0}{2}\leq t\leq t_0 \\
0 & t_0\leq t
\end{cases}$$
and then the height
$$x(t)=\begin{cases}
+mgt^2 & 0\leq t<\frac{t_0}{2}\\
-mg(t- \frac{t_0}{2})t+L& \frac{t_0}{2}\leq t\leq t_0 \\
L & t_0\leq t
\end{cases}$$
A point to mention is that this isn't the only function that would fit the conditions, there would be an infinite number of possible force functions. But they would still have an average force of $mg$.
hopefully that helps
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How does light from distant galaxies reach us? Why doesn’t a ray of light which is emitted by a distant galaxy, say about a thousand light years away, die down in between?
I mean, how do the light rays from different galaxies so far away reach us in this day and age, when they were emitted so long ago? Why aren’t they dampened and lost altogether? What drives the light rays to travel humongous distances WITHOUT being dampened or being lost altogether?
Now, how to explain this if constant momentum and energy are what drives a photon through space (according to https://physics.stackexchange.com/a/667784/311056)?
| As other answers tried to explain, it's not about what mechanism would prevent photons from getting lost but finding a mechanism that would remove sufficient photons so we couldn't observe them any longer.
according to our understanding of fundamental physics:
mass less objects travel with the "speed of light"
According to relativity, everything with zero mass always travels with the (local) speed of light. There is nothing "driving" these objects. It is a fundamental property of spacetime. Mass is "resistance to acceleration", without mass, there is no resistance and objects move with the "maximum allowed velocity".
energy is conserved
Once you have created a photon, you need to find a mechanism to get rid of it again.
Photons do not decay, so you need other entities to interact with and dissipate / absorb them, cf. e.g. the argument concerning dust and other molecules.
If you cannot find an explanation why we wouldn't see light from billions of light years away then that is your explanation why we do.
| {
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Where should the reference point be considered during torque measurement? This is an extremely silly and wierd question.
https://en.wikipedia.org/wiki/Varignon%27s_theorem_(mechanics)
While reading about Varignon's Theorem in wikipedia I noticed this sentence,
"If many concurrent forces are acting on a body, then the algebraic sum of torques of all the forces about a point in the plane of the forces is equal to the torque of their resultant about the same point."
Well what does "a point" specify here? Can it be any point even outside the object which we are dealing with?
Another thing, lets say we are dealing with 2 concurrent forces about a point C and the two forces are acting on A. Now according to the above theorem shouldn't the above scenario be looked as
$$F_1
*AC+F_2*AC=F*CC=F*0=0?$$
Perhaps I misunderstood the theorem. As for the 1st question, I have never thought of any reference point in case of rotational motion other than the center of mass of the object and mostly dealt with uniform shapes.
| Not so silly. Toppling of a cylinder on a block may help set the stage. A torque is a pair of anti-parallel forces of equal magnitude with a sideways displacement applied to an object.
Suppose you have a pulley supported by a fixed axis. Suppose a bunch of forces are applied to the pulley, causing it to spin. The fixed axis supplies a reaction force to each applied force. So each applied force is half of a pair of forces that apply a torque to the pulley.
$$\tau = F d_{perp}$$
Since the magnitudes are equal, you can calculate with either force. All the reaction forces are applied to the same place. So it is easy to ignore the reaction forces and focus on the applied forces.
In cases like this, we can speak of each applied force as generating a torque about the axis.
| {
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Can we bend a light ray into any closed loop? Suppose we have a medium with varying refractive index and a source of light inside that medium emitting rays. Is it possible to bend the ray into any closed loop?
As the medium has varying refractive index, is it possible?
And if possible, how will it look like if anyone stand in the path of the ray?
| Ok so here are my thoughts. I wanted to construct something like a clock where each circular sector is made of a different material. The incoming angle of the light ray with respect to the boundary of the first to the second layer is $\alpha$ (see image). To make things easy I wanted to choose the second material and circular section in such a way, that the incoming angle in the second refraction is still $\alpha$. If the refractive index of the second material is bigger than the index of the first material, this means that the circular sector of the second material must be smaller $\alpha - \epsilon$. Since the sum of the angles in a triangle is $\pi$ we find that the outgoing angle $\beta$ must be
\begin{equation}
\beta + \pi/2 + \pi/2 - \alpha + \alpha - \epsilon = \pi \quad \Rightarrow \quad \beta = \epsilon.
\end{equation}
Hence we must choose two materials with
\begin{equation}
\frac{\sin \alpha}{\sin \beta} = \frac{\sin \alpha}{\sin \epsilon} = \frac{n_2}{n_1}
\end{equation}
For the next circular sector, we proceed in the exact same way. The sector has the angle $\alpha - 2 \epsilon$ and the incoming angle is still $\alpha$, thus the ratio $\frac{n_3}{n_2}$ is equal to $\frac{n_2}{n_1}$. This process can be repeated until we got around the entire circle. Important is that we have to choose $\alpha$ and $\epsilon$ in a way that we get around the entire circle, i.e.
\begin{equation}
\sum_n (\alpha - n \epsilon) \ge 2\pi.
\end{equation}
For instance choose $\alpha = 30^\circ$ and $\epsilon = 1^\circ$.
This does not entirely answer your question though, as the loop is not closed. The distance to the center of the circle gets larger and larger with every refraction. In the ideal case $\epsilon = 0$ you would have a closed-loop but for that, you would need a material with an infinite refractive index. So for a setup like this, we have shown that it is not possible to construct a closed loop. And even if I did not show it I predict that it is not possible at all (but I am happy to be convinced of the opposite).
Also notice, that even if you are happy with a loop that is not closed, you can not build such a device in real life as you run out of materials with a high enough refractive index. For $\alpha = 30^\circ$ and $\epsilon = 1^\circ$ you would already require $n_2/n_1 = 29$ which is impossible to achieve in real life (as far as I know).
| {
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Group velocity (open vs periodic boundary conditions) I'm trying to understand the meaning of the group velocity for Bloch electrons given by
$$
\mathbf{v}=\frac{1}{\hbar}\frac{\partial E(\mathbf{k})}{\partial \mathbf{k}}
$$
where $E(\mathbf{k})$ is the energy of the band, and $\mathbf{k}$ is the crystal momentum. So defined, this velocity is only applicable for periodic boundary conditions.
However, in the thermodynamic limit, the bulk properties should be independent of the boundary conditions, so $\mathbf{v}$ should describe the velocity of bulk eigenstates of the Hamiltonian. Generally, $\mathbf{v}$ depends on $\mathbf{k}$, and under open boundary conditions $\mathbf{k}$ is not a good quantum number. This confuses me:
What is the relation between the Bloch velocity and the (local) velocity of the bulk electrons under open boundary conditions?
| The short answer is, in the thermodynamic limit, k could always be defined as a good quantum number for bulk states despite of different boundary conditions.
This can be shown by considering a finite potential of several equally distributed wells and infinite high walls as boundary.
The solution of wave function is then planewave-like and has some phase shift if we move from one well to another. And, we can see that at the thermodynamic limit (# of wells -> infinite), the phase shift is nothing but exp(-ikR), where the definition of k is more complecated than the perodic case. However the difference eventually goes to zero if we take the thermodynamic limit. Therefore we can still construct wave packets with these newly defined "wave vectors", and thus the group veolocity.
| {
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Mixed symmetry of rank $3$ tensor I have rank 3 tensor $T_{ijk}$ with following properties:
$T_{ijk}=T_{jik}$
$T_{ijk}=-T_{kji}$
Is it true that there is the only one tensor of rank 3 with those properties and it is $T_{ijk}=0$.
I'm starting from the following
$T_{ijk}=-T_{kij}=-T_{ikj}=T_{jki}=T_{kji}=-T_{ijk}$
$\Rightarrow 2T_{ijk}=0$
$\Rightarrow T_{ijk}=0$
Where am I making a mistake? Because the result is strange enough. Does this mean that from the point of view of Young diagrams, the hook is not interesting?
| Your work seems fine. This little fact is sometimes known as the $S_3$-lemma, and it can be stated in a very general way: if $S$ is a non-empty set, $G$ is a (say, abelian) group without elements or order $2$, and $T: S\times S \times S \to G$ is symmetric in the first pair, and skew-symmetric on the second pair, then $T =0$.
$$T(x,y,z) = T(y,x,z)=-T(y,z,x)=-T(z,y,x) = T(z,x,y) = T(x,z,y) = -T(x,y,z).$$
The fact that tensors represent multilinear maps is irrelevant here. If $G$ is non-abelian, same proof works replacing minus signs with inverses (this is what would be understood as skew-symmetry).
| {
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Is pulling a string instantaneous at both ends? Why is it or isn't it? This is a question that has bothered me for quite some time but I don't have a clue where to start in researching it.
Let's say we have a string which is arbitrarily long, light enough to not take much effort to pull but rigid enough that it would not break or stretch. If I now stand at one end of the string and pull on it, does the other end move instantaneously? If so, could we use such a method to send a message faster than light? I'm not referring to the speed of the rope itself, but to the time between pulling one end and the other end moving accordingly.
My knowledge of FTL messaging/travel (basically gen-ed physics and the movie Interstellar) tells me the answer is no, and there's a whole lot more going on at the atomic level, but I don't know exactly what forces are at play.
| First of all, you can't even imagine the force required to pull a string of length equal to or more than $3×10^8\ \text{m}$. Now, imagine you have the force to pull it, but it is nearly impossible to make a string of that rigidity. It can't withstand that enormous force.
Now, imagine you have both source of force and an extremely rigid string. Now you are pulling or pushing the string, but the fact is the force doesn't show up at the end instantaneously. Force travels through the rope as a form of longitudinal wave like this,
Because when you apply force on one end, it propagates through the collision of atoms, the atom at the back pushes the front one, and thus force propagates.
Note: When you take a small string and pull it, it seems like the force reaches the other end instantaneously, this happens because it is small and thus force reach the other end very fast, but for long rope, particularly this big ($3×10^8 \ \text{m}$), it is not instantaneous.
| {
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Conditions for a planet to become spherical I have a question, from which size/mass will a body in space adopt a spherical shape? Over 500 kilometers wide and/or 1/4 the mass of Pluto?
Something like that, I always had this doubt.
| The minimum size has been called the potato radius, as anything too small will look more like a potato than a sphere. The potato radius depends on composition, hereafter assumed uniform. Eq. (9) (ibid.) gives the radius as proportional to $\rho^{-1}\sqrt{\sigma_y/G}$, with $\rho$ the density and $\sigma_y$ the compressive strength. This is to be expected from dimensional analysis. The authors estimate a $300 \text{km}$ value for representative values of $\rho,\,\sigma_y$.
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Why do rain drops fall with a constant velocity? While reading my physics book. I came across a line that says that:
Rain drop falls with a constant velocity because the weight(which is the force of gravity acting on body) of the drop is balanced by the sum of the buoyant force and force due to friction(or viscosity )of air. Thus the net force on the drop is zero so it falls down with a constant velocity.
I was not satisfied by the explanation So I searched the internet which too had similar explanations:
The falling drop increases speed until the resistance of the air equals the pull of gravity, at which point the drop begins to fall at a constant speed, its terminal velocity.
My confusion regarding the matter is that if the net force acting on a body (here the rain drop) is zero then it should remain suspended in air rather than falling towards the earth. So how come the rain drop keeps falling when net force acting on it becomes zero? How the air resistance and other forces stops the rain drop from acquiring accelerated downward motion?
| This is Newton's first law: if the force vanishes the velocity is constant. Constant but not necessarily zero.
The resistive force increase as long as speed increases. When it is equal to the gravitational force the speed no longer increases and the total force vanishes leading to constant speed.
| {
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Confusion with Impulse and Work I get Momentum and Impulse as well as Work and energy but struggle when it comes to connecting the two ideas. I understand that an objects KE can change without its momentum changing, Like in a inelastic collision were the KE of the system decreases but the amount of matter moving to left with some speed - the amount of matter moving to the right with some speed will be equal before and after the collision. Hence Momentum is Conserved. My confusion starts with the question is an impulse of a force on an object always accompanied by some work done on the object by that force?
| Impusle is a property defined as: $$\vec{I} = \int \sum \vec{F}.dt$$
and by virtue of this definition, it need not always be accompanied by work.
Consider an example of you standing on a skateboard and pushing on a wall setting yourself in motion away from the wall.
In this case, the force from the wall on your hands moves through no displacement; the force is always located at the interface between the wall and your hands. Hence, no work is done on the system of you and the skateboard. (The resulting kinetic energy however comes from the potential energy stored in your body).
The impulse-momentum theorem will suggest that: $$\Delta \vec p_{tot} = \vec I = m \Delta \vec v = \int \vec F_{\text{wall}}.dt$$ where $\vec F_{\text{wall}}$ is the force exerted by the wall on your hands. Hence, the impulse does not result in a work.
Hope this helps.
| {
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Quantum Mechanical Interpretation of Rutherford Experiment Ernest Rutherford performed the gold foil experiment; alpha particles were fired at a gold foil and the alpha particles were scattered. This result disproved Thomson's plum pudding model of atoms.
This got me wondering, how does QM fit into this picture? How do we use the wave nature of alpha particles to explain what is going on in this experiment? When does the wavefunction collapse? Did the wavefunction spread out across the entire space prior to measurement? If so, doesn't it mean that the particles were not "fired" but rather just under the influence of the potentials of the gold atoms?
P.S. My understanding of the physical interpretation of QM is all over the place so my question might not even be valid in the first place. If so, sorry for that.
| There are two parts in the question. First, given potential, how to find the scattering amplitudes as a function of scattering angle. Second, how to obtain the scattering potential in the first place.
The answer to the first question is twofold again. Normally, yes you need QM to compute scattering amplitudes correctly. However Rutherford did not do that and just did the calculation using classical scattering theory for a Coulombic potential. But there he was helped by one of those rare lucky coinsidences that helped researchers when they needed to leap the gap. Namely, the scattering amplitudes for Coulombic field coinside for classical and quantum theories. This is a feature of Coulombic potential that makes it special in this sense.
Second, the scattering potential. This is where QM was necessary. Classical theory could not (and still cannot) accommodate for a point-like concentration of positive charge in the center of an atom without compromising its long-term stability. Rutherford reached this conclusion with his classical scattering calculation, which he was lucky is still valid in QM domain, and the rest is history.
| {
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What makes the (non-abelian) strong interaction so special that it leads to confinement? The strong interaction has a coupling constant of $\alpha_s(91GeV)\approx 0.1$ whereas the weak interaction has a much lower coupling constant $\alpha_w \approx 10^{-6}$. Both theories are non-abelian gauge theories, the strong interaction is based on SU(3) gauge symmetry, whereas the electroweak interaction is based on $U(1)\times SU(2)$ gauge symmetry.
What makes the strong interaction so special that it leads to confinement, whereas for the electroweak interaction it is not the case? It is certainly related with the $\beta$-function of the corresponding interaction, but why is the $\beta$-function of the electroweak interaction positive and the $\beta$-function of the strong interaction negative? Actually, I am not very familiar with use of renormalisation group arguments, so I would prefer a not too formal answer based on essentially on physics arguments.
| Just to state the result for the beta-function associated to QCD
\begin{equation}
\beta = -\frac{g^2}{32 \pi^2} \left(\frac{11}{3}N_c - \frac{2}{3}N_f \right)
\end{equation}
in which $N_c$ is the amount of colours and $N_f$ is the amount of flavours.
Essentially, both terms boil down to antiscreening and screening respectively.
A single quark can be surrounded by quark-antiquark pairs which tend to screen it from effects of the environment (much like in QED with electron-positron pairs).
However, the important piece is now the antiscreening effect due to the self-interaction of the gluons (because of the non-Abelian gauge symmetry). This tends to make the quark more susceptible to its environment.
Filling in the constants gives a negative beta-function. In other words, if we probe quarks at higher energies, the coupling constant decreases sufficiently such that we may regard them as being free (eg. a quark-gluon plasma). At small energies, the coupling constant becomes enormous and quarks are effectively bound together into hadrons - confinement.
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Eccentricity of planets based on distance from Sun Are the orbits of inner-solar system planets more circular than outer planets? Or is it the other way around? What's the reason for this? We were taught in our high school Physics class that outer planets had more circular orbits, but some sources online and even on SE state otherwise.
| The eccentricity of a planet's orbit is a measure of how 'circular' it is.
According to this website (14th line down) - there is no clear pattern in the eccentricities of the planets.
| {
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Inverse Laplacian I have seen the following operator somewhere in a paper on cosmology
$$
\frac{\partial_i \partial_j}{\nabla^2} - \frac{1}{3} \delta_{ij}.
$$
What is the definition of the inverse Laplacian? What is meant by this misleading notation? Is this the inverse Laplacian? If yes, what is then $\frac{1}{\nabla^2}$?
| Every Laplacian can be inverted using its Green's function. If we have
$$\nabla^2V = \rho$$
the inverse is simply
$$V(x) = \left(\nabla^2\right)^{-1} (\rho) = -\frac{1}{4\pi}\int \frac{\rho(x')}{|x-x'|} \text{d}^3x'$$
and that's what's meant by $\frac{1}{\nabla^2}$, at least in the places where I've seen it.
| {
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Sustaining a current in conductor placed in external electric field Consider the following arrangement-
We have a conducting sphere and a positively charged infinite sheet on the left. The field creates induced charges and the net electric field inside the conductor is zero after a very short time. During this short time, there is a current in the conductor as electrons as dragged opposite to the external electric field.
My teacher says that if we want to sustain this brief current we should connect it with a conducting wire making a closed loop. The electrons will flow anticlockwise giving a steady current, which I think is wrong.
When we connect the conducting wire, the sphere and the wire become one complete metal and after a short time again, there will be induced charges in this big metal and electrostatic condition will be reached. The potential inside will be the same everywhere so how will the current flow in a closed loop?
| In order to sustain a current, one needs to sustain a potential difference between two parts of the conductor. If we have a closed loop, the charges eventually redistribute so as to screen the potential and the current ceases. However, if, for e.g., there is a battery in the loop, which absorbs charges on one side and injects them on the other while maintaining the potential difference, the equilibrium is never reached. One could also drive the current by changing the magnetic field through the loop, i.e., using the Faraday effect.
| {
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Graph for Coulomb Force vs $1/r$ My teacher told me that the graph for the coulomb force $F$ vs $1/r$ where $r$ is the distance between the 2 charges should be parabolic but I can't seem to understand why. I am aware that equations of the form $y^2=4ax$ are parabolic but why should $F$ vs $1/r$ graph be parabolic?
| $$F = k_e\frac{q.Q}{r^2} \rightarrow (\frac{1}{r})^2 = \frac{F}{k_eq.Q} \rightarrow x^2 = 4ay$$
where $x = 1/r$ and $4a = 1/(k_e.qQ)$
Hope this helps.
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Clausius inequality leading to absurd result Background: After deriving Clausius inequality, the author of this book derives the following relation:
Consider the cycle shown in the figure in which leg $A \rightarrow B$ is irreversible. In the equation
$$
0>\oint\frac{\mathrm{d}Q}{T}=\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d}Q}{T}+ \int_{B \operatorname{rev}}^{A} \frac{\mathrm{d}Q}{T}
$$
the second term on the right-hand side of this equation is given by $S(A)-S(B)$ because it is taken over a reversible path. When we move this quantity to the left-hand side, we find that
$$
S(B)-S(A)>\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}.
$$
Thus the difference in entropy between the points is greater than the integral of $\mathrm{d} Q / T$ over an irreversible change.
Problem: Entropy is a state function so $\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}= {\Delta} S. $ By the inequality derived we have ${\Delta} S>{\Delta} S$ which is absurd.
| Considering the result $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}<S(B)-S(A)$$ for an infinitesimal path, we get $$\mathrm{d}S\ge\frac{\mathrm{d}Q}{T}$$ where the equality holds only for a reversible process (by the definition of entropy).
This means that in your expression $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T},$$ $\frac{\mathrm{d} Q}{T}$ is not equal to $\mathrm{d}S$ because the process is irreversible. Instead, you have $\frac{\mathrm{d} Q}{T}<\mathrm{d}S$ and so $$\int_{A \operatorname{irrev}}^{B} \frac{\mathrm{d} Q}{T}<\int_{A \operatorname{irrev}}^{B} \mathrm{d}S=S(B)-S(A)$$ which is the original result.
| {
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How much longer could Titanic have stayed afloat if it gotten rid of its anchor and chain right after hitting the iceberg? I am wondering how much longer the RMS Titanic could have stayed afloat if the crew had allowed the ship's anchor and anchor chain to fall to the bottom of the ocean immediately after the ship had hit the iceberg. (I am not even sure if a ship's anchor chain can be unfastened from a ship, but let's just say for the sake of this question that it can be unfastened.)
The combined weight of Titanic's anchor and anchor chain was approximately 116 tons according to this Wikipedia article:
"...In 1911, the company manufactured the anchors and chain for the ocean liner RMS Titanic. The largest of the anchors weighed 15.5 tons and on completion was drawn through the streets of Netherton on a wagon drawn by 20 shire horses.[15] The chain and fittings for the anchors weighed around 100 tons..."
https://en.wikipedia.org/wiki/N._Hingley_%26_Sons_Ltd
Since the Titanic went down bow first, and the anchor and chain was located in the bow section, immediately getting rid of 116 tons in the bow section would have increased the time it had stayed afloat before it sank.
| A detailed question to this answer will, as so often, require a detailed numerical simulation or, better, a scaled-down experiment. But being the gung-ho physicists we are on this site here, we can do a simple guesstimate: According to a quick internet search, the total mass of the titanic is somewhere in the ballpark of $5\times 10^7$kg, so the mass you are talking about is only a small $10^{-3}$ fraction. So to first order one might guess that this would delay the sinking by such an amount: With the total time between hitting the iceberg and sinking being about 2.5 hours (says wikipedia), $10^{-3}$ of that is of order 10 seconds. Rather not a lot.
More detailed studies have undoubtedly been done and lead to estimates like the one quoted on, again, wikipedia, of 7 tons of water flooding the titanic per second. That means the mass of anchor and chain is made up entirely within just a few seconds also, again giving an unfortunately also negligible delay estimate of some 16 seconds.
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Are there any physical symmetries that do not form a group? I'm trying to understand the role of symmetry in physics. The overall derivation seems pretty intuitive, reasonable and elegant, but one point I'm curious about is the identification of physical symmetries with groups. It does not seem a priori necessary that the transformations under which a mathematical object remains invariant obey the group axioms of associativity and invertibility. Identity seems necessary, since leaving the object unchanged obviously should not produce a change in the invariant, but it is at least conceivable that invariants can exist under non-associative or non-invertible transformations. Are there any examples of such invariants?
| Symmetries act on objects. For a symmetry to be nonassociative, it would have to be the case that (doing A and then doing B) and then doing C was different from doing A and then (doing B and then doing C). I don't think that's possible.
There is a previous question about invertibility with some good answers. In short, there are noninvertible analogs of symmetries, but they often aren't called symmetries (i.e., "symmetry" is taken to imply invertibility by definition).
| {
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Could a non-photon massless particle travel at a speed other than $c$? The speed of light is given as $c=\frac{1}{\sqrt{ε_0μ_0}}$ which is in terms of the electric and magnetic constants.
Hypothetically, another massless particle could exist which does not interact with the electronic or magnetic fields. Postulate that a "foo" field and a "bar" field exist which were mathematically analogous to the electric and magnetic fields, the "foobaron" particle might have a speed given by $c_{fb}=\frac{1}{\sqrt{f_0b_0}}$, derived in exactly the same way that $c$ is derived, but using different, independent fields.
If such a particle were to exist, must $c=c_{fb}$? Why?
| c is the speed of cause and effect. See Do we know why there is a speed limit in our universe?
The speed of light is a special case of this. If you move a point charge, the "news" that it has moved will spread out at this speed. That is, changes to the electric and magnetic fields from the charge will spread out at this speed. Forces from the charge acting on another charge will not change until the "news" arrives. See In what medium are non-mechanical waves a disturbance? The aether? for more.
It is possible for waves and such to travel at slower speeds. For example sound does. This is because it is a pressure wave. Atoms bump into other atoms and push them. It takes a while for atoms to accelerate. The pushing arises from electric fields in the atoms' electrons. The rate at which neighboring atoms begin to feel the push spreads out at the speed of light.
| {
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Why $E$ is neglected at large and small $r$ of quantum harmonic oscillator? In obtaining radial solution of quantum oscillator why E is neglected?
Radial equation:
Resource: nouredine zettili.
| One does this because the original differential equation is overly difficult to solve without first examining the boundary conditions.
The point here is that you are initially only approximately solving the TISE in the regimes of small $r$ or large $r$. This is where the boundary conditions are, so you want to get the correct behaviours at these boundaries. For finite energy, the dominant term is, in one case, the $1/r^2$ term (assuming $\ell\ne 0$), and in the other case the $r^2$ term.
Once you have the correct behaviour near $r=0$ and at $r\to\infty$, the strategy is then to interpolate between small $r$ and large $r$ using a polynomial.
Thus you start with small $r$ (or Eq.(6.86)), and then go to large $r$ (or Eq.(6.87)) and this gives you a solution of the form
$$
U(r)\sim r^{\ell+1} e^{-\lambda r^2}f(r)\, , \tag{1}
$$
for some unknown polynomial $f$. This way, you are guaranteed that the $U(r)$ you find will satisfy the correct boundary conditions.
You then plug (1) into the full TISE to get a differential equation for the unknown function $f$, which can be solved (after some gymnastics) by identification with a standard differential equation.
This is similar to the 1d harmonic oscillator, except there is no condition the solution for small $r$: in 3d you need this solution to go to $0$ smoothly since the system cannot have “negative values of $r$”. In the 1d case, there is a condition at $r\to\infty$ and one often starts there to get the right behaviour, then interpolate back to finite $r$s.
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What is the physical meaning of "correlation length"? I am studying phase transitions right now and trying to understand the physical meaning of the concept correlation length. I saw the equations but I still couldn't quite wrap my head around the physical meaning of it. Like is it the length of the correlation between neighbouring atoms or what? And what does it mean for correlation length to be large or small? System is considered more organized when correlation length is large, right? And what is the meaning of correlation length being infinite, like when it is infinite, can we say that all atoms in the system are correlated with each other?
| Here's an example using a crystal. The same concept applies to many other physical systems.
Consider an ideal crystal. If you know the position of atom at a, then you can predict the position of atom b no matter how far away it is. But real materials are not ideal crystals. There are inclusions, faults, vibrations ... so the atoms are not at their ideal location. If I know the position of a, I probably know the position of its nearest neighbor to an excellent degree of certainty. The next nearest neighbor ... slightly less certainty. The atom a kilometer away, not at all. The length over which the positions can be reliably predicted is the correlation length.
| {
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Why proper time is a measure of space? Recently I've been trying to learn General and Special Relativity by myself. There is an specific thing I do not understand perfectly, proper time in the metric of the space-time.
Take the case of an empty space-time:
$$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2$$
where $\tau$ is the proper time of an object.
I don't understand when $c^2 \mathrm d \tau^2$ is used as $ds^2$. Why is this possible? It is related to the worldline traced by an object? Could you calculate the line integral in order to find the length of the path the object? This line integral would give the path length of a geodesic?
Could you get the speed of an object by doing this (correct me if I'm wrong):
$$-c^2 \mathrm d \tau^2 = -c^2 \mathrm d t^2 + \mathrm d x^2 + \mathrm d y^2 + \mathrm d z^2 \rightarrow \\ -c^2 \left( \frac{\mathrm d \tau}{\mathrm d t} \right)^2 = -c^2 + \left( \frac{\mathrm d x}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d y}{\mathrm d t} \right)^2 + \left( \frac{\mathrm d z}{\mathrm d t} \right)^2 $$
If you assume that: $\dot{x}^2 + \dot{y}^2 + \dot{z}^2 = ||v||^2$, then by rearranging a little bit the equation, you could get the velocity of an object.
| In relativity, notions like "space" and "time" depend on what coordinate system you use. However, there are some physical quantities which all observers will agree on. Say there is a clock which travels along some trajectory, from one point $x_i$ in spacetime to another point $x_f$. (Points in spacetime have an invariant meaning, but the way we label them will depend on the coordinate system we use, i.e. the spacetime point $x$ could be written in some coordinates as $(x^0, \vec{x})$ or in some other coordinates as $({x^0}', \vec{x}')$.)
All observers will agree on what time the clock reads when it starts its journey and what time it reads when it ends its journey. The time elapsed will be given by the proper time integral from the start point to the end point
$$
\Delta \tau = \int_{x_i}^{x_f} d \tau.
$$
This is what the spacetime metric gives you: an invariant measure of how much proper time passes when a clock travels along a path.
| {
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How does Upthrust/buoyant force act on an object? I came across this question where, In a container there's water at the bottom , kerosene on top and an ice cube floating between them and I was asked to calculate the ratio of height of cube in ice to that in water:
Now till now my intuition for up thrust was that there needs to be some fluid below the object to give an upward perpendicular force.As simple as this:
Therefore for this question I thought there should be no upthrust from kerosene but I was surprised when I saw the free body diagram in the solution:
And therefore now I wish to understand:
*
*What's wrong with my understanding of upthrust?
*Where is the upthrust from kerosene actually coming from and How does upthrust acts on a body in general?
| This diagram was used in an answer to this question Why does a fluid push upward on a body fully or partially submerged in it?
The spring is first at its natural length, then released, showing that there is an up-thrust on the piston due to the height difference (and weight) of the liquid.
In a similar way, even though there is no Kerosene underneath the cube, the weight of it (transmitted through the water) adds to the upward pressure and force pressing on the underneath the cube. On the top side of the cube, there is only the weight of a reduced depth of Kerosene pressing down.
So the result is that there is also an up-thrust due to the weight of the Kerosene displaced, as well as due to the weight of water displaced, as described by Archimede's principle.
| {
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Would somebody feel a magnetic field if they are travelling at the same velocity as a charge? I am little bit curious about how magnetic fields are being generated when a charge moves.
I want to check if somebody travelling along with a charged particle, would that person experience a magnetic field?
How are magnetic fields really generated?
| Magnetic and electric fields are just aspects of what's called electromagnetism. The electromagnetic field is what's unambiguously there, and so is the electromagnetic force on the person.
Whether you see the electromagnetic field as just an electric field or also having a magnetic component depends on your velocity.
In your example, the moving particle creates a larger electric field than a particle at rest would, and it also generates a magnetic field. Both act on the person, resulting in some net force F.
OTOH, as linear motion is only relative, you're also allowed to look at the problem from a frame where both are at rest. In such a frame, there is no magnetic field, and the electric field a bit smaller, resulting in the same force F.
So, the magnetic field is "created" by a moving electrostatic field. But keep in mind that both are a bit arbitrary, depending on the frame you choose to look at things. Only their combination, the electromagnetic field, is unambiguous.
| {
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A question regarding commutators in quantum mechanics I propose the following thought experiment:
Suppose we have a beam of identically prepared electrons that is splits into two. The first goes through detector A that detects the $x+y$ where $x$ is the coordinate along x direction and $y$ is the coordinate along the $y$ direction. Then, we measure the difference of the momenta of the electrons in the $x$ and $y$ directions i.e. $p_{x}-p_{y}$. Then, according to the postulates of quantum mechanics, we can measure the both quantities to arbitrary precision since
$$ [x+y, p_{x}-p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$
The second beam of electrons is subjected to a similar measurement by a detector B but this time we measure $x-y$ and then measure the sum of momenta i.e. $p_{x}+p_{y}$. Then, again we can measure $x-y$ and $p_{x}+p_{y}$ to arbitrary precision because
$$ [x-y, p_{x}+p_{y}]=[x,p_{x}]-[y,p_{y}]=0$$
Then, adding the results of the measurements we have $(x+y)+(x-y)=2x$ and then $(p_{x}-p_{y})+(p_{x}-p_{y})=2p_{x}$. Both of which, $x$ and $p_{x}$ can be measured to arbitrary precision thus violating the uncertainty principle.
If on the other hand, we carry out this experiment and find that we are not able to measure the above quantities to arbitrary precision then it follows that the postulates of quantum mechanics do not correctly predict the outcome of the experiment in the sense that the commutator vanishes but we can't measure the quantities to arbitrary precision.
Does this mean that the postulates of quantum mechanics are inconsistent? (I certainly don't hope so!)
| It would be far simpler to just directly measure $x$ of your first beam and $p_x$ of the second beam.
Both of which, $x$ and $p_x$ can be measured to arbitrary precision thus violating the uncertainty principle.
There is no violation of the uncertainty principle. If you have an unlimited supply of identically prepared systems you can measure to arbitrary precision (in principle). The uncertainty principle limits what you can simultaneously measure on a single system.
| {
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Does deceleration require energy? Consider an apple falling from a tree and striking the ground.
The ground decelerates the apple once it hits it, but the force is not applied over any "distance" - it is experiencing the force when it is in contact with the ground - so no work is done, yet there is a change in momentum, what is going on with the energy here?
| The force is applied over a distance. Both ground and apple are deformed by the impact force. If the apple falls from a large height you can easily see the effect of deformation. In general, the amount of deformation depends on the nature of the objects colliding, their hardness and elasticity. If they are both elastic the deformation is reversible and in the second part of collision the force is accelerating the colliding objects.
| {
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Intuition for Stress and the Cauchy Stress Tensor I'm struggling to get an intuitive understanding of what exactly Stress is, particularly the "direction" associated with it.
In the case of a 1 dimensional bar with just uniaxial loading, the way stress was explained to me was just $\pm\frac{F}{A}$ where $F$ is the force applied to either end, A is the cross sectional area, and the sign refers to tension or compression. This explanation is fine as a formula, but I don't see how it relates to "internal forces".
I've found other sources explaining it with an "imaginary cut" through the material, ignoring one side of the cut, and imposing equilibrium on the other piece. Why can either side be "ignored"? If stress is the internal force per unit of surface, why doesn't the neglected part contribute to the stress? (after all, the neglected part shares the surface).
In the more general case using the Stress Tensor,
$$T=\begin{bmatrix}
\sigma_{xx} & \tau_{xy} & \tau_{xz}\\
\tau_{yx} & \sigma_{yy} & \tau_{yz}\\
\tau_{zx} & \tau_{zy} & \sigma_{zz}\\
\end{bmatrix}$$
Do each of the components describe the stress on the surfaces of some infinitesimal volume? If so, which faces do they describe (there are 2 faces normal to each direction)- is it the sum of the stresses on each face?
Any insight on these questions would be greatly appreciated, thanks for reading
| We could start from the Cauchy postulate where we take an arbitrary plane defined by a normal vector $\hat{n}$ and with a finite area $\Delta S$, and compute the traction in that plane $\Delta R$ (here I am neglecting the couples $\Delta M$ since they should vanish in the classical case).
If we take the limit when the size of the are goes to zero we end up with the definition of the traction at a particular point (in index notation)
$$t_i^{(\hat{n})} = \sigma_{ji} n_j\, .$$
What this is telling us is that the effect that we are seeing (traction) is the projection of a tensor $\sigma_{ji}$. That means that the internal "forces" are characterized by this tensor and we probe them via a projection, computing tractions.
Then, we can interpret the components of the stress tensor using the plane over which we are projecting and the direction associated with the given traction vector. As shown in the following figure from Wikipedia's article con Cauchy stress.
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Normalization constant of blackbody radiation We have the Boltzmann-Distribution: $$~~~~~~~~~~~~~ P_n = N(T) \cdot \exp \left(- \frac {n \cdot h \cdot \omega }{2 \pi \cdot k_B \cdot T }\right) ~~~ , ~n \in \mathbb N_0 ~~~$$ ($n$ = number of photons)
The sum over all the possibilitys $P_n$ must be one.
Now we have to show with the geometric series $ ~\sum_{n=0} ^ \infty x^n = \frac 1 {1-x } ~~~, |x|<1 ~~~$ that normalization constant is $N(T)=1-\exp\left(-\frac {h \omega }{2\pi \cdot k_B \cdot T}\right) $
What I did:
$~~ \sum P_n = 1 = \sum N(T) \cdot \exp\left(- \frac {n \cdot h \cdot \omega }{2 \pi \cdot k_B \cdot T }\right) = N(T) \cdot \sum \exp\left(- \frac {n \cdot h \cdot \omega }{2 \pi \cdot k_B \cdot T }\right) $
Now I'm confused, the base of exp(...) is e>1, so we can't use the geometric series on this point.
| From your
$$1 = \sum N(T) \cdot exp(- \frac {n \cdot h \cdot \omega }{2 \pi \cdot k_B \cdot T })$$
$$\frac{1}{N(t)} = \sum\ exp(- \frac {n \cdot h \cdot \omega }{2 \pi \cdot k_B \cdot T })$$
Letting $x = exp(- \frac { h \cdot \omega }{2 \pi \cdot k_B \cdot T })$
and using the identity in your question should give the result
$$\frac{1}{N(t)} = \frac{1}{1-exp(-\frac {h \omega }{2\pi \cdot k_B \cdot T})}$$
then do the reciprocal of both sides.
The geometric identity could be used because $e^{-y}= \frac{1}{e^y}$ is less than one for any positive $y$
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How much does a radiator at 25˚C affect a room that is 20˚C? This is probably not a very precise question but I'm just trying to get a rough idea how much effect a radiator set at such a low temperature would have on a room. Let's assume that the number of radiators and size of the radiators is generally appropriate for the size of the room. Make up your own approximate values for these.
I want to get some idea whether it has some effect or is basically negligible in a real world application and if it does affect the room's temperature, how long would it take to do that.
| If the temperature of the radiator is higher than the temperature of the room then there will be a flow of heat from the radiator to the room. The rate at which heat flows into the room will be proportional to the difference between the radiator temperature and the room temperature.
If the room is hermetically sealed and perfectly insulated then the temperature of the room and everything in it will increase until it reaches the same temperature as the radiator. The time that this takes (or, more exactly, the time it takes for the room temperature to get close to the radiator temperature, since exact equality will take an infinite length of time) will depend on many factors, including the size of the room, the size and material of the radiator, the contents of the room. This case is dealt with in detail in Gert's answer. Note that in no circumstances can the radiator heat the room to a temperature that is higher than the radiator temperature - this would violate the second law of thermodynamics.
However, in reality it is likely that the room will not be hermetically sealed and perfectly insulated, and the temperature of the outside environment is likely to be no higher than the room temperature. In this case, as the room temperature rises, there will be a flow of heat energy from the room to the outside environment. The steady state room temperature will now be the temperature at which the rate of heat flow from the room to the outside equals the rate of heat flow from the radiator into the room - this will be lower than the radiator temperature but greater than the outside temperature.
So the best we can say for a real room is that given enough time the radiator will heat the room to some temperature between $20^o$C and $25^o$C, depending upon how well sealed and insulated the room is.
| {
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Eigenfunctions of an Harmonic Oscillator perturbed with an Electric Field Knowing that a particle of mass $m$ and electric charge $q$ is under an uni dimensional harmonic potential of frequency $\omega$ perturbed with and electric field $\vec{E}= E_f \hat{x} $, $\hspace{0.2cm}$ I would like to find the eigenfunctions and e
eigenvalues of $\hat{H} = \frac{p^2}{2m} + \frac{1}{2} m \omega^2 x^2 - q E_f x $.
I know that I can find the eigenvalues by noticing:
\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \frac{1}{2} m \omega^2 \bigg( x^2 - \frac{2 q E_f}{m\omega²}x + \frac{q^2 E_f^2}{m^2 \omega^4}x - \frac{q^2 E_f^2}{m^2 \omega^4} \bigg) = E \psi \Leftrightarrow
\end{equation}
\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2} + \frac{1}{2} m \omega^2 \bigg( x - \frac{q E_f}{m \omega^2} \bigg)^2 = \bigg(E + \frac{q^2 E_f^2}{2 m \omega^2}\bigg)\psi
\end{equation}
So I use the change of variables: $x' = x - \frac{q E_f}{m \omega^2}$.
Using this, I get that the new eigenvalues are of $\hat{H}$ are: $E = \hbar \omega \bigg( n+ \frac{1}{2} \bigg) - \frac{q^2 E_f^2}{m \omega^2}$.
My question is: If I use the change of variables I found in the expression of the eigenfunctions of the harmonic oscillator, do I get the orthonormal eigenspace for this specific potential and how I could prove it?
Furthermore I would like to have a physical explanation to understand why this trick works?
Thanks in advance
|
My question is: If I use the change of variables I found in the expression of the eigenfunctions of the harmonic oscillator, do I get the orthonormal eigenspace for this specific potential and how I could prove it?
Yes. First, we can see it directly: if $\psi_n(x)$ and $\psi_m(x)$ are eigenfunction of the original harmonic oscillator (before the shift in variables) then we know they are orthogonal i.e. $$\int_{-\infty}^{\infty}dx \psi^*_m(x)\psi_n(x) = \delta_{m,n}$$ so a constant shift $\tilde{\psi}_n(x)=\psi_n(x-x_0)$ will leave this relationship intact
$$\int_{-\infty}^{\infty}dx \tilde{\psi}^*_m(x)\tilde{\psi}_n(x) = \int_{-\infty}^{\infty}dx \psi^*_m(x-x_0)\psi_n(x-x_0) = \int_{-\infty}^{\infty}dx \psi^*_m(x)\psi_n(x) = \delta_{m,n}$$
because it is just a shift in the variable of integration, equally amont both function.
Another way to see it is a bit more formally, without resorting to the representation of the eigenfunction in position space. The shift is done by a unitary transformation $|\tilde{n} \rangle = T_{x_0} |n\rangle$ where $T_{x_0}=\exp(-ipx_0/\hbar)$ is the generator of translations. Since this is unitary $T_{x_0}T^{\dagger}_{x_0} = 1 = T^{\dagger}_{x_0}T_{x_0}$ then it doesn't change inner product between states $\langle \tilde{m} | \tilde{n} \rangle = \langle m | T^{\dagger}_{x_0}T_{x_0}|n\rangle = \langle m | n \rangle$. This is useful to remember as it is applicable to all such unitary transformations, regardless of the specific problem at hand.
Furthermore I would like to have a physical explanation to understand why this trick works?
The reason is similar to the classic case of a spring under gravity, for example. The equilibrium point of the the harmonic oscillator was shifted due to the force. So now, instead of having $x=0$ as the equilibrium, you have $x=x_0 = qE_f/m\omega^2$ as the equilibrium, but the particle will oscillate about this new equilibrium just as it has oscaillated about $x=0$ before hand. As long as we don't add terms with higher orders of $x$ than $x^2$, the harmonic nature of the potential is preserved, just around a new point (or, if we add a term $\propto x^2$, a new frequency). So this shift in variables just translates the problem to its true equilibrium.
| {
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Confusion on molecular dynamics (MD) simulation units leading to absurd acceleration values I am trying to code up a very simple MD simulation in order to learn more about it. I am using the Leonard-Jones potential, expressed as $ V=4\epsilon (\frac{\sigma}{r}^{12}-\frac{\sigma}{r}^{6}) $
The pairwise force is therefore $\frac{1}{r}\frac{dV}{dr}=\frac{24\epsilon}{r}((\frac{\sigma}{r})^6-2(\frac{\sigma}{r})^{12})$
I want to use parameter values as listed in Li's publication:
http://li.mit.edu/A/Papers/05/Li05-2.8.pdf
However, when I work through the math, I end up with unreasonable values as shown in the following calculation:
Suppose two argon atoms are separated by 4 angstroms. The ambient temperature is 300K.
In Joules, $\epsilon=119.8k_B T=119.8\cdot 1.3806\cdot 10^{-23} \frac{J}{K} \cdot 300K=4.9619\cdot 10^{-19} J$.
The pairwise force is then $4.9619\cdot 10^{-19}J * \frac{24}{4\cdot 10^{-10} m} ((\frac{3.405Å}{4.0Å})^6-2(\frac{3.405Å}{4.0Å})^{12}) = 2.26\frac{J}{m}=2.71\cdot 10^{-9}N$.
The mass of one a.m.u. in kg is $1.6605\cdot 10^{-27} kg$.
$f=\frac{m}{a}\implies a=\frac{f}{m}=\frac{2.71\cdot 10^{-9} N}{39.948*1.6605\cdot 10^{-27} kg}=4.0854\cdot 10^{16}\frac{m}{s^2}$
My simulation has units of angstroms, so I usually convert the acceleration to angstroms.
$ a=4.0854 \cdot 10^{26} \frac{Å}{s^2}$.
This number is absurd. Even with a timestep of 1ps, the particles will fly out of the bounding box in just a few simulation steps.
Where did I go wrong?
EDIT: Thank you to @Samson for providing the correct pairwise force equation, $\vec{F_{IJ}=\frac{48\epsilon}{\sigma^2}[(\frac{\sigma}{r_{IJ}})^{14}-0.5\cdot(\frac{\sigma}{r_{IJ}})^{8}]}\vec{r_{IJ}}$. Here, $\sigma$ has units of angstroms, $\epsilon$ units of Joules, $r_{IJ}$ units of angstroms. $\vec{r_{IJ}}$ is the vector difference of the pair's positions, i.e. $\vec{r_I}-\vec{r_J}$.
If you use that equation, and handle the units correctly, you get acceleration on the order of $10^{25} \frac{Å}{s^2}$. Regardless of what integrator you use, the total displacement after one timestep will be proportional to the timestep squared. If the timestep is $10^{-13}$ seconds or so, the order of magnitude of displacement in each step will be less than one angstrom, which is an agreeable number to work with. In my implementation, I expressed the time-step as multiples of picoseconds (in some cases, a fraction of a picosecond) and did the unit conversions on paper. This avoids using extreme floating-point numbers and leads to better numerical stability.
| I am not familiar with the exact molecular dynamics model you are using, but I have done some simulations with Newton acceleration on mass or Coulomb acceleration on charges. It can be striking what is embedded in those simple accelerations when time-evolved. If the model just simulates Newton's a = G/r2 attraction, we can get Kepler's laws from time evolving its contents. From Coulomb's a = K*q1q2/r2 we get Gauss's law and some others. So, one way is you can perhaps change it to a simpler equation at first and see the results.
| {
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Is it the term "telescope" the same as a "detector"? For example, in this reference, MITO: muon telescope they use the term telescope but clearly the "telescope" is a muon detection system. And they also talk about angular resolution, angular aperture, etc. So my question is focused in, is the term telescope the same as a particle detector? and if a particle detector can be described by the properties of an optical-telescope(talking about optical geometry, diffraction, angular resolution, ect.)?
| It is in the sense of definition 2 of wesbster dictionary
any of various tubular magnifying optical instruments
where the detector is used to record rays,(instead of the eye or film) directional vectors, composed of particles (muons in this case)
| {
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Why do the electron and positron creation operators anti-commute? I am learning QFT and is baffled by a minor problem.
The electron and the positron should be distinguishable, as they have different charges. So why do their creation operators anti-commute? They should commute with each other, just like the creation operators of the electron and the proton commute with each other.
| As far as I remember, you can choose creation operators of different fermions to commute or to anti-commute, both choices should provide correct results if used consistently.
| {
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Why should a dipole have zero net charge? Why can a dipole not have two unequal charges separated by a distance? Is there any significance for the dipole being defined as electrically neutral?
| If we have several charges distributed in space, then the total potential is
$$
\phi = k\sum_{n = 1}^{N} \frac{q_{n}}{\left|\mathrm{r} - \mathrm{r}_{n}\right|}
$$
If charges are concentrated in a narrow region of space (narrow compared to the distance to the observer and, if there are EM waves, to the wavelength) then we can expand the fraction linearly assigning $r_{n} = r_{0} + \Delta_{n}$:
$$
\frac{1}{\left|\mathrm{r} - \mathrm{r}_{n}\right|} = \frac{1}{\sqrt{(\mathrm{r} - \mathrm{r}_{0} - \mathrm{\Delta}_{n})^2}} \approx \frac{1}{\sqrt{(\mathrm{r} - \mathrm{r}_0)^2 - 2 (\mathrm{r} - \mathrm{r}_0) \cdot \mathrm{\Delta}_{n}}} \approx
\\
\frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \frac{1}{1 - \frac{(\mathrm{r} - \mathrm{r}_0) \cdot \mathrm{\Delta}_{n}}{(\mathrm{r} - \mathrm{r}_0)^2}} \approx \frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} + \frac{1}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \mathrm{\Delta}_{n}
$$
For the first approximation I neglected quadratic term of $\Delta$, for the second i used Taylor series for the square root, for the third I expanded the fraction in Taylor series.
Substituting it into the equation for the potential, we get
$$
\phi \approx \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|}\sum_{n = 1}^{N} q_{n} + \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \sum_{n = 1}^{N} q_{n} \mathrm{\Delta}_{n} = \frac{k Q }{\left|\mathrm{r} - \mathrm{r}_{0}\right|} + \frac{k}{\left|\mathrm{r} - \mathrm{r}_{0}\right|^2}\frac{\left(\mathrm{r} - \mathrm{r}_{0}\right)}{\left|\mathrm{r} - \mathrm{r}_{0}\right|} \cdot \mathrm{d}
$$
Here $Q$ is the total charge, and $d$ is the dipole moment. As you see, the first term decays as $1/r$ and the second one as $1/r^2$. If the first one is present (i. e. $Q\ne0$) then the second term can be, usually, neglected when we are far away form the system of charges.
| {
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Why bother buying efficient lights if you are already heating your house? Assume I live in a location where at any time of day and any time of year, I need to heat my house. Assume further that I have a room with no windows. In this case, does it make sense for me to buy efficient light bulbs, considering that any inefficiency in converting electricity to visible light simply leads to more heat being added to the room, which in turn, results in less heat being output by the heater to maintain constant room temperature.
Although these are somewhat idealized conditions, I don't think they are too far off from being realistic. For example, say you live near the arctic circle, it might be smart not to have many windows due to heat loss, and it seems reasonable that in such a climate, heating will be required at all times of the day and year. Assuming I haven't missed something, it seems to me, somewhat unintuitively, that buying efficient light bulbs is not a logical thing to do. Is this the case?
| With more energy efficient lights you can also have brighter ones while still staying within safe watt limits. Less energy used for heat means you can get more light and still not come close to your wiring's tolerance. It's even more important when you have a lighting fixture with it's own maximum wattage. This is due to danger from both the heat generated by the bulb as well as the power draw.
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Euclidean space to Minkowski spacetime Can you continuously deform (i.e., shrink, twist, stretch, etc. in any way without tearing) four-dimensional Euclidean space to make it four-dimensional Minkowski spacetime?
| Of course!
It is called Wick rotation. You have to change the time coordinate $t$ to $-it$, which changes the sign of the action in the partition function of a physical system. It is a basic tool is statistical physics. The partition function then changes from $\exp(-iS)$ to $\exp(S)$. As I'm writing its rather backwards, typically the action of a physical system is given in Lorentzian signature and the rotation changes it to Euclidean. Then you perform calculations/simulations in Euclidean signature and perform an analytic continuation and rotate the solution back to Lorentzian.
There are some caveats in particular cases, but this is how it works.
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Weight of Diamagnetic Objects in Lenz Effect I've seen the simple experiment of dropping a neodymium sphere down through a copper or aluminum tube where it falls at less than g. The "Lenz Effect"
My question is:
Has there ever been an experiment when the NeoD sphere is tied to a string attached to a weight scale? Does the sphere weigh less at the midpoint inside of the tube than it does moving outside of it?
| Someone may have done such an experiment, but it would be messy. It would be a lot easier to measure how long it takes the magnet to fall through the tube to get the same result.
Assuming the magnet falls with a uniform acceleration, since $$y=\frac{1}{2}at^2 \\
\rightarrow a=\frac{2y}{t^2}$$ Since you know the length of the tube $y$, and if you measure the time the magnet takes to fall through the tube $t$, then you can calculate the net acceleration of the magnet.
From this you can calculate the (net) force applied on the magnet, which is simply $$F_N=ma=\frac{2my}{t^2}$$ where $m$ is its mass. This force will obviously be smaller than the magnet's weight force.
| {
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Is the Integer Quantum Hall Effect a distinct phase of matter? In the Landau classification scheme, phases of matter differ in terms of symmetry. However, we know of many instances where this classification scheme does not apply. I have often heard topological insulators and even the different plateaus in the integer quantum hall effect referred to as distinct phases of matter. What does this mean? Do these states really qualify as genuine phases? I am perfectly aware that there may not be consensus on this issue, but I'd like to know the different viewpoints on this issue.
| Yes, integer quantum Hall phases are distinct phases of matter with invertible topological order. Any phase of matter can be distinguished from another by the presence of a phase transition, where a smooth change in parameter(s) leads to a discontinuous change in some observable quantity. In the case of the integer quantum Hall phases, different phases can be distinguished by their transverse electromagnetic and thermal responses. In spite of the fact that integer quantum Hall phases contain no non-trivial topological excitations, they are topological in that they cannot be smoothly connected to unentangled product states by local unitary transformations (that don't grow in size with the system size).
There may not be a consensus on details like whether IQH phases are long range or short range entangled, and I am not sure if everyone agrees about whether or not they are SPT phases (without any symmetries), but I don't think there is a lack of agreement about the fact that they are distinct phases of matter.
| {
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Klein-Gordon Field Quantization and Bose-Einstein Statistics in Peskin & Schroeder I am trying to understand how Klein-Gordon particles obey Bose-Einstein statistics from Peskin & Schroeder's QFT textbook (page no. 22). The excerpt is given below:
From this passage it is clear to me that the two-particle states: $a_\textbf{p}^\dagger a_\textbf{q}^\dagger|0\rangle, a_\textbf{q}^\dagger a_\textbf{p}^\dagger|0\rangle$ are equal due to the commutation relation: $[a_\textbf{q}^\dagger, a_\textbf{p}^\dagger] = 0$ derived earlier in the chapter. They represent to a two-particle system of total energy $\omega_\textbf{p} + \omega_\textbf{q}$ and total momenta $\textbf{p} + \textbf{q}$. But the following sentence is not clear to me:
Moreover, a single mode $\textbf{p}$ can contain arbitrarily many particles...
Also, I don't see how the statements of this paragraph fit with the fact that for a two-particle Bosonic (or, Fermionic) system, if we swap the particles the wave function does not (or, does) change. Therefore, I couldn't follow the argument of why Klein-Gordon particles obey Bose-Einstein statistics. Could you please help me to understand this?
| $\newcommand{\Ket}[1]{\left|#1\right>}$
In the Klein-Gordon field, a creation operator $a_p^\dagger$ represents the creation of a single particle in mode $p$. The statement that a single mode can contain arbitrarily particles is simply a statement that, just like the quantum harmonic oscillator, $(a_p^\dagger)^n \Ket{0}$ does not vanish and gives a unique state for every $n$ (representing a state with $n$ particles in mode $p$).
| {
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Theoretical minimum temperature required to melt any material Reading about this (New material has a higher melting point than any known substance) got me curious.
Given a pressure level (like 1 atm) and a sufficiently hot temperature, I have the intuition that no material stays solid, and turns to plasma if hot enough.
So here's the question: According to modern physics models, what is the lowest known temperature beyond which we can guarantee that any material will be past its melting point? We can consider an arbitrary material sample being heated under isobar conditions at 1 bar.
Can we in theory make a material that remains solid at 1 bar and 4500K? 6000K? 20000K?
| The outer part of a neutron star is considered solid and its temperature can reach $10^6$ K. This is probably the highest temperature that a solid can reach.
| {
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How often is a non-coordinate and non-orthonormal basis used in GR? I wrote a program that takes as input the basis vectors if electing to use an orthonormal basis, or metric components if using the coordinate basis, and outputs non-zero Christoffel symbols and components of Riemann, Ricci, and Einstein tensors, as well as the Ricci scalar.
I could include functionality to support a non-coordinate and non-orthonormal basis, but I don’t want to waste my time if that’s something that no one ever uses in GR. I know that I don’t know enough about GR yet to make a call on this, so I’m asking you all!
| I’ll take the above comment by Sebastian Riese as my answer:
This is typically called vierbein, and used in many ways (e.g. to simplify certain calculations, in Einstein-Cartan theory or for defining spin on curved space times). More info: https://en.wikipedia.org/wiki/Tetrad_formalism
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Scattered radiation factor for diagnostic
I need the confirmation, is it if we put survey meter at 270 degree, does the reading of scattered radiation is same with 90 degree. In my opinion, the reading would be same as the distance is still same (other factor like kVp, Feild size, thickness are constant). or do we need to considered back scattered radiation as it deflect 180 degree.
my explaination is like this-the reading of survey meter would be like 90°, this is because the location and distance of survey meter from x-ray tube is similar with 90°. However, if the survey metre is placed close to the x-ray tube's anode, the reading may be slightly greater than cathode side. The anode is the component of the x-ray tube that produces x-rays.
| If you need confirmation that scattering at 90° is the same as scattering at 270°, you should measure it.
In one of the first experiments on polarizing Mott scattering of electrons, in 1927, Cox and collaborators built an angular detector like the one you are describing here. They included “redundant” detectors at 90° and 270° as an experimental control, expecting those detectors to give identical results, but they did not. After a heroic experimental effort, Cox et al. discovered that their “control” measurement was well-behaved if they used electrons from a thermal source, but not if they used electrons from a beta emitter.
The significance of this observation wasn’t understood until 1958, in a literature review by Grodzins. Cox et al. had discovered that electrons from beta sources are already polarized when they’re emitted; this is a violation of “parity symmetry” which occurs only in weak interactions. There is a lovely explanation in Allan Franklin’s book “Are there really neutrinos?” Franklin calls the Cox et al. experiment the “non-discovery of parity non-conservation.”
You are right to suspect that the 90° and 270° measurements should be identical. That’s why you measure them. If they come out different, that difference will show up somewhere else in your experiment as well.
| {
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Existence of Ground State of Dirac equation In chapter four of Ryder, the author showed that there exists a ground state $|0\rangle$ for the Kelin-Gordon equation, just like the case of the linear harmonic oscillator. However, I was not able to extend this to show the existence of a ground state for the Dirac field equation.
So, how does one prove the existence of a ground state for the Dirac field equation?
| The Dirac equation spectrum is not bounded from below, hence there is no ground state. To augment this deficiency Dirac postulated that the spectrum of negative energy states was already filled (the so called Dirac Sea). He then predicted the antimatter state (positron) as a hole In this sea.
| {
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How can we make the energy levels of an one-dimensional infinite square well potential equispaced? We know that the energy values for an one-dimensional infinite square well potential is given by $$E_n = \frac{n^2{\pi}^2{\hbar}^2}{2ma^2}$$ where $a$ is the width of the well. Now, as we can see that the difference in energy of two consecutive levels is given by $$E_{n+1} - E_n = (2n+1)E_1.$$
In my PhD interview, it was asked that how can we make these energy levels equispaced?
I know that the quantum harmonic oscillator energy levels are equispaced, but how can we make the levels equispaced for one-dimensional infinite square well potential? Is this related to Bohr Correspondence Principle somehow?
| They just wanted you to use the formula for the energy and adapt it very likely. This means if you know the formula for $E_n$, you can see that a cuadratic dependence on $n$ will not lead to equispaced energy levels. So which parameter could you play with so that $E_n$ does not scale cuadratically with $n$?
You might think of $m$ or $a$ but, let us say the particle you are given has a fixed mass, so the only thing remaining is $a$, you could then require that $a$ is not constant and you make it such that $a^2 \propto n$ then
$$\Delta E_n = E_{n+1}-E_n = \frac{\pi^2\hbar^2}{2m}.$$
This turns out is achieved by the potential not being a square well, but a parabolic well, i.e. a harmonic oscillator.
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If space is a vacuum, how do stars form? According to what I have read, stars are formed due to the accumulation of gas and dust, which collapses due to gravity and starts to form stars. But then, if space is a vacuum, what is that gas that gets accumulated?
| As people like to think,Vacuum is space that is devoid of matter like air etc. Space is not strictly a vacuum at all places, but it is better vacuum then we obtain in laboratories. when we say space is vacuum it means it doesn't have air to breathe. But space still has matter. You are right universe including stars was formed when gas called solar nebula and dust collapsed. Space is very vast. It has dust particles, gases like helium or hydrogen, rocks, asteroids, planets, galaxies, oort cloud, orion, black holes, satellites etc. the reason we call space a vacuum is that the atmospheric pressure of outer space is very low. Vacuums are regions of low pressure.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why are time-ordered Greens functions equal to retarded Greens functions at zero temperature? When I calculate a photon polarization diagram:
I get the same answer:
*
*If I calculate it in equilibrium (retarded Greens functions) with finite chemical potential, in the limit of zero temperature, or
*If I calculate it as a scattering amplitude (time-ordered Greens functions) in a particle bath $|\psi\rangle = \prod_{|\vec{p}|<p_F}c^\dagger_{\vec{p}}|0\rangle$.
The equality follows from a weird cancellation from the Fermi-Dirac distribution at zero temperature. Is there a theorem or some argument that this should happen in general (i.e. for more general correlation functions)?
| This is generally not the case. The reason is that, in addition to zero temperature, you are averaging over the vacuum state with zero particles, which is killed by annihilation operators.
Let us take time ordered Green's function at zero temperature:
$$
G^t(t,t')=\langle 0|T[c(t)c^\dagger(t')]|0\rangle=\\
\langle 0|c(t)c^\dagger(t')|0\rangle\theta(t-t') \pm \langle 0|c^\dagger(t')c(t)|0\rangle\theta(t'-t)
$$
and the retarded Green's function
$$
G^r(t,t')=\langle 0|[c(t)c^\dagger(t')]_\pm|0\rangle\theta(t-t')=\\
\langle 0|c(t)c^\dagger(t')|0\rangle\theta(t-t') \pm \langle 0|c^\dagger(t')c(t)|0\rangle\theta(t-t')
$$
If
$$ c(t)|0\rangle=0$$ the second terms in both equations are zero and the two functions are identical.
However, e.g., if the ground state is a Fermi sea filled up to its chemical potential the result is different.
| {
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Squared Summation of Terms using Einstein's summation convention In working with QFT and Maxwell's equations, terms such as:$$\left(\partial_\mu\,A^\mu\right)^{2}$$ often appear. Since I am new to this, I am not sure of the expansion. That is, is it 4 terms squared or is it 4 squard terms:
$$\left(\partial_0 A^0\right)^{2} + \left(\partial_1 A^1\right)^{2} +\left(\partial_2 A^{2}\right)^{2} +\left(\partial_3 A^3\right)^{2}$$Or,
$$\left(\partial_0 A^0 + \partial_1 A^1 +\partial_2 A^{2} +\partial_3 A^3\right)^{2}$$
| Reference : Squaring the E&M (Maxwell) field strength tensor.
Looking at equation (06) of my answer in above link we realize that this square comes from the expression
\begin{equation}
\left(\partial^{\mu}A_{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)\left(\partial_{\nu}A^{\nu}\right)\boldsymbol{=}\left(\partial_{\mu}A^{\mu}\right)^2
\tag{06}\label{06}
\end{equation}
so is the square of a sum and not the sum of squares.
| {
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RMS velocity of a gas vs RMS velocity of a gas molecule, which is a more appropriate term? I'm a bit confused by the terminology. Is it the RMS velocity of a gas, or the RMS velocity of the gas molecules or of a gas molecule?
Similarly, is it the mean velocity of a gas, or the mean velocity of the gas molecules or of a gas molecule?
| It is the molecules which have the velocity not the gas as a whole.
The RMS velocity of the molecules in a gas (mean taken at an instant of time) should be the same as the RMS velocity of a gas molecule (mean over a "long" period of time) but given the qualification (and uncertainty) regarding the length of time go for The RMS velocity of the molecules in a gas.
| {
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Sound proofing: mass-spring-mass or mass-mass-spring? I am trying to improve the sound proofing of a metal box. The box is made of steel of thickness $0.8 \;\mathrm{mm}$. I have additional sheets of steel with $3 \;\mathrm{mm}$ thickness for reinforcement. For vibration reduction, I have access to some bitumen mats (anti-vibration mats for damping cars) and some alubutyl (butyl rubber with an aluminum foil on top).
What would be the best way to improve the insulation properties of the box?
(A) Glue both steel plates together $(0.8 \;\mathrm{mm}+3.0 \;\mathrm{mm})$, then add bitumen/alubutyl to reduce vibrancy.
(B) Put the anti-vibration mat right into the middle to achieve kind of a mass-spring system.
The bitumen mats are pretty stiff and thus will probably be not sufficient to act like a damping spring. The alubutyl however is more like gooey grubber and might be better for attenuating the mechanical vibration. On the other hand: It is only $2 \;\mathrm{mm}$ thick.
My take on that is that (A) might be a more rigid construction (if glued tightly) while (B) might be better for blocking mechanical vibration to travel from the inside to the outside.
The noise source I want to isolate (computer hard drive) adds both structure-borne and air-borne resonance in/to the case.
I am looking forward to your suggestions.
| Here are some general rules of thumb:
Before you start designing your sound blocker, you need to know the spectrum of sounds you wish to block. This will determine which wall design will do the trick. For example, if the noise spectrum consists primarily of high frequencies (above, say $3000 \;\text{Hz}$), then two sheets of steel with a stiff rubber layer in between will be the best. If the spectrum consists primarily of low frequencies (say, $500 \;\text{Hz}$ and below), then sheets of steel with soft rubber sandwiched in between will be the best. For very low frequencies $100 \;\text{Hz}$ and below), lots of mass (many steel sheets with soft rubber in between) will be needed. For very high frequencies (say $6000 \ Hz$ and above), a single sheet of steel and a single sheet of stiff rubber will be the best.
| {
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Why only the wavelength and speed of refracted light traveling inside a transparent material changes and not its frequency? When monochromatic light waves travel from one medium to another the frequency never changes.
A transition to a denser medium will result of a slow down of the propagation speed v of the light wave and its wavelength λ but not of its frequency f. Photons still travel at c speed from one atom to the next through the vacuum space between the atoms of the medium.
$$u=\lambda f$$
Should not both λ and f proportionally decrease to match the slower v speed?
Why f does not change what is the physical explanation?
Also, if f does not change and since no transparent material is perfect and there will be apart of reflection also some absorption, how light absorption, lost energy, is then justified by the equation
$$\mathrm{E}=\mathrm{h} \mathrm{f}$$
if f remains unchanged?
| Due to continuity if one wavelength of light crosses a boundary, one must emerge on the other side. Hence the number of wavelengths crossing the boundary per second is equal to the number emerging per second. So the frequency of the incident and refracted wave must be the same.
For your second question, energy is lost by whole photons being absorbed/reflected, this is because they are a quanta of EM radiation, so cannot be subdivided.
| {
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Polarization of natural light for double slit interference Last week I performed Young's double slit experiment using a laser. As expected, I obtained an interference pattern as predicted by Fraunhofer theory (enveloped by the 1 slit diffraction curve). Then, I added two polarizers with perpendicular axes in front of each slit. Since the light coming from the slits had now different polarizations, I no longer noticed the interference pattern (only the diffraction from each separate slit).
It is generally known that in order to perform Young's double slit experiment using natural light, we should first make the natural light "more" coherent by placing a single slit before the double slit. However, from my understanding, natural light is unpolarized. So now my question is, why does natural light produce an interference pattern if the incoming light in unpolarized? Shouldn't the light from the 2 slits not interfere because the polarizations are different just like in the laser experiment? Does the first slit polarize the light? Or is it that the light exiting the 2 slits is synchronously unpolarized (the electric field vectors are always on the same line) hence creating a pattern?
| Both polarizations destructively and constructively interfere at the same points, meaning their sum also shows the diffraction pattern. This is why unpolarized light can be used in Young's experiment with no problem.
However, by letting a different polarization to pass in each slit, you force a situation where each polarization doesn't have two sources to interfere. (meaning there is a sum of two diffraction patterns from a single slit that can't interfere)
| {
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Projectile motion given angular projection When a body is projected with a certain angle from the ground it returns with the same angle and speed with which it was projected. What if the the body was projected from some height, maybe a building. Will it still reach the ground with same speed and angle?
| No, because once the particle goes below the level of projection, the speed increases, due to conservation of energy. Also the angle of flight, given by $\tan\phi=\frac{\dot{y}}{\dot{x}}$, increases because $\dot{y}$ increases while $\dot{x}$ remains constant.
| {
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Question about dark matter/energy and other dimensions According to drummer and lyricist Henrik Ohlsson, the title Dark Matter Dimensions refers to the "appreciation and acknowledgement of the unseen worlds and dimensions, because without the existence of these unseen forces our physical universe would never be able to exist". So could dark matter be regular matter, just in other dimensions?
| Dark matter simply refers to matter that does not emit light. In other words it does not interact with the electromagnetic field. We have never been able to observe it directly, since our telescopes can only see incoming light from the universe.
But we can infer it's existence due to the fact that gravitational lensing due to galaxies is a lot stronger than we'd expect. We tried to account for all of the mass of a galaxy due to stars, planets, gases, etc, but there is still a lot of mass unaccounted for. Without this unaccounted for mass, we would not see gravitational lensing as strongly as we do. So there is some sort of additional mass that we cannot see that is in galaxies.
It is unlikely that dark matter is matter in another dimension. When we observed black hole mergers, the resulting gravitational waves had the same energy when they reached us as we would predict if space only had 3 dimensions. It is unlikely that there are any 'hidden' dimensions, as we have only observed gravity to act in 3.
| {
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What is the current status of the convergence of the post-Newtonian approximation? In the very well written article by C. Will, On the Unreasonable Effectiveness of the post-Newtonian Approximation in Gravitational Physics, he states:
The one question that remains open is the nature of the post-Newtonian sequence; we still do not know if it converges, diverges or is asymptotic. Despite this, it has proven to be remarkably effective.
What is the current status of the convergence of the post-Newtonian sequence? Are there indications that it is possibly asymptotic?
| The question about PN sequence is raiser mathematical one, nor physical or philosophical. It has a good answer in the paper "Newtonian and post-Newtonian approximations are asymptotic to general relativity" by T. Futamase and Bernard, F. Schutz, Phys. Rev. D 28, 2363 – Published 15 November 1983. The Abstract to this paper is clearly stated that the PN approximations "are genuine asymptotic approximations to general relativity". Also this question discussed in many papers such as
Blanchet, L., “Gravitational Radiation from Post-Newtonian Sources and Inspiralling Compact Binaries”, Living Rev. Relativity, 9, lrr-2006-4, (2006). URL (cited on 3 August 2006): http://www.livingreviews.org/lrr-2006-4. 2.2, 3
"The Post-Newtonian Approximation for Relativistic Compact Binaries", by Toshifumi Futamase&Yousuke Itoh, Living Reviews in Relativity, 12 Mar 2007, 10(1):2
"Accuracy of the post-Newtonian approximation: Optimal asymptotic expansion for quasicircular, extreme-mass ratio inspirals" by Nicolás Yunes and Emanuele Berti, Phys. Rev. D 77, 124006 – Published 5 June 2008; Erratum Phys. Rev. D 83, 109901 (2011).
| {
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The change of mechanical into electromagnetic waves and vice versa I know that sound is a type of mechanical wave, so the human eardrum changes mechanical energy into electronic energy (impulses) so the information may be processed by the brain.
Question: As satellites transfer info by electromagnetic waves that are also electric signals, then can we change these mechanical waves into electromagnetic waves and vice versa?
| Sound waves have frequencies from 40 to 20000 Hz. At a sped of 340 m/s this leads to wabelengths of 8m down to 17 mm. Light waves in this frequency range have wavelengths between 7500 km and 15 km. Usually you are detecting at a much shorter length scale, at which these waves look like AC. This is also why I don't care about the length difference between my speaker cables.
| {
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Question about indices and matrix This is essentially a trivial question, which can be answer probably immediately, but i have this doubt anyway.
If, say, $$\Lambda^{a}_{b} = \begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}$$
$$\Lambda^{b}_{a} = \begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}^T or \begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}^{-1}$$ ??
For example, suppose $$T^{c'd'}=T^{cd}\Lambda^{c'}_{c}\Lambda^{d'}_{d} = = \begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}T\begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}^T or \begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}T\begin{pmatrix}
f & -fc\\
-fc & f
\end{pmatrix}^{-1} $$
?
| Here is the rule that always works: always multiply matrices by keeping the indices which are being summed over touching each other. The upper/lower placement doesn't matter. Only the right-left.
The standard matrix multiplication rule is
$$
(AB)_{ik} = A_{ij} B_{jk}.
$$
Notice how the the two instances index that is being summed over, $j$, are adjacent to each other. This is what corresponds to normal matrix multiplication.
When it comes to raising/lowering, this doesn't affect the matrix multiplication rule.
However, if you have an expression like
$$
A_{ij} B_{kj}
$$
where the two instances of $j$ aren't adjacent, then you must take the transpose so that they are. For instance, the above expression corresponds to $$A B^T.$$
So for instance, say you have
$$
\eta^{\mu \nu} = \eta_{\mu \nu} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}
$$
and
$$
\Lambda^\mu_{\;\; \nu} = \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}
$$
which implies
\begin{align}
T^{\mu' \nu'} &= \Lambda^{\mu'}_{\;\; \mu} \Lambda^{\nu'}_{\;\; \nu} T^{\mu \nu} \\
&= \Lambda^{\mu'}_{\;\; \mu} T^{\mu \nu} \Lambda^{\nu'}_{\;\; \nu} \\
&= \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix} T \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix}^T.
\end{align}
Also notice that
\begin{align}
\Lambda_{\mu}^{\;\;\nu} &= \eta_{\mu \mu'} \eta^{\nu \nu'} \Lambda^{\mu'}_{\;\; \nu'} \\
&= \eta_{\mu \mu'} \Lambda^{\mu'}_{\;\; \nu'} \eta^{\nu' \nu} \\
&= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \gamma & \gamma v \\ \gamma v & \gamma \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \\
&= \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v & \gamma \end{pmatrix}
\end{align}
| {
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Selection Rules using Group Theory I was learning about the applications of Group Theory and one important application is looking at the selection rules in a weak EM field. We essentially want to see whether the matrix element $\langle i|H'|f\rangle$ vanishes due to the reasons of symmetry. Here $i$ denotes the initial state, $f$ denotes the final state and $H'$ denotes the perturbed part of the hamiltonian. Dresselhaus (link: http://web.mit.edu/course/6/6.734j/www/group-full02.pdf, page 139, equation 7.29) states that the perturbed Hamiltonian can be expanded in terms of the irreducible representations of the group of the Hamiltonian.
How can we have a right to do this as the perturbed hamiltonian is not necessarily in the group of the unperturbed Hamiltonian? I understand that by Maschke's theorem we can expand any representation of group $G$ by its irreducible representations. But can an arbitrary reducible representation of a group $H$, be expanded by the irreducible representations of $G$? Furthermore, even though the irreps (irreducible representations) of $G$ form on orthogonal basis, how can we make sure that the perturbed Hamiltonian spans a space that has the same dimensionality as the irreps of the Group of the Hamiltonian?
|
We essentially want to see whether the matrix element <i|H'|f> vanishes due to the reasons of symmetry
How can we have a right to do this as the perturbed hamiltonian is not necessarily in the group of the unperturbed Hamiltonian ?
The states "|i>" and "|f>" are typically taken to be states of the unperturbed Hamiltonian. Otherwise you can't get very far, in general.
For example, if i and f are single particle atomic states (states for a spherically symmetric potential) and the perturbation is the
$\vec{r} \cdot \vec E$, you can expand i and f in terms of $Y_{lm}$ functions. You can also write $\vec r \cdot \vec E$ as a linear combo of the $Y_{1m}$ functions. Then you get selected rules pretty straightforwardly (Clebsch Gordan coefficients and whatnot).
Here, it is the spherically symmetric potential of the unperturbed Hamiltonian that allows us to use the $Y_{lm}$ functions to expand the unperturbed eigenstates in a helpful way.
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Can we measure $10^{-12}\ \mathrm{N}$ force? I would be interested to measure a very small force, say in the order of $10^{-12}\ \mathrm{N}$? Is this possible? What equipment is needed?
My setup
Assume that I have a relatively heavy machine say between 5-10 kg that I want to measure if it produces this thrust, which according to calculations should be of this feeble magnitude. But (according to the predictions) this should be periodic, with a frequency of about 200Hz and it should last for about a quarter of the time period. I should also mention that this apparatus is expected to vibrate (a little), since inside a disk is supposed to be rotated at about 12k rpm.
My research
I have read about torsion balance as a possible method. I am also thinking about some piezo-electric crystals. Would be feasible?
What piezoelectric cells would be recommended? I read that the Atomic Force Microscopy devices are also implemented using piezoelectric materials.
| Not to answer your question completely but on the AFM point- we can model the AFM cantilever as a spring in contact mode. The spring constant can be ~0.2 N/m or lower.
Now you can, without much work get z-direction sensitivities of 0.1 nm without too much trouble, taking into account experimental noise.
Very roughly speaking this gives a resolution of 0.02 nN or 20 pN. So, not far off for measuring forces on a tiny cantilever. However, for a mass that size I'm not sure it's possible.
| {
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What does the term "condense" mean in the physics literature? When reading the physics literature, we often see the term "condensate".
Some examples:
*
*in the string net model (Wen, Levin), one will say the string "condensate".
*in QCD, people talk about the quark-gluon condensate.
*In stat mech, one talks about the bose einstein condensate.
*People will talk about 'monopole condensation' to describe confinement transition.
My understanding is that in some cases the ground state acquires an interesting non-zero expectation value of "something" after the phase transition, but even then it's not clear.
What do condensed matter physicists mean when they say something "condenses"? (a mathematical definition in terms of operators would help).
If there are other examples of condensation, I'd be interested to learn more.
| Condense could probably mean :-
*
*to become denser or more compact or concentrated.
*Change of state of matter.
Condensate could mean :-
*
*The liquid phase produced by the condensation of steam or any other gas.
*The product of a chemical condensation reaction, other than water.
*Natural-gas condensate, in the natural gas industry.
Now there is more to it in deep physics like :-
*
*Bose-Einstein condensate is a state of matter that occurs when a set of atoms is cooled almost to absolute zero in which a statistical description of the positions of the atoms implies that they physically overlap each other and in effect form a single atom.
*Fermionic condensate is basically a pairing of electrons (which are fermions) to produce a condensate is a crucial feature of superconductivity, so a fermionic condensate would give us crucial insights into the mechanisms behind superconductivity, as well as superfluidity.
*Gluon condensate, a non-perturbative property of the QCD vacuum.
*Color-glass condensate (CGC) is a type of matter theorized to exist in atomic nuclei when they collide at near the speed of light.
| {
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Why doesn't a parallel circuit violate conservation of energy? Let's imagine a hypothetical circuit where there are a large number of wires placed in parallel to each other, hooked up to a simple power source.
We know that voltage at each wire would be equal $V_{total}=V_1=V_2=...=V_n$ where $n$ approaches a large number; and that each wire is of some arbitrary constant length.
Next, assume that at the start of each wire there is a single charge of $+1C$, in each wire placed in parallel.
Since work done on a charge is $W=VQ$; where $W=$ work done, thus we apply the same voltage to each charge in each wire placed in parallel.
Since the voltage across each wire would be the same (say, $Resistance$ is ineligible, but $\neq0$) the work done would be same.
Additionally, we know $W=\vec{F}.\vec{s}$; Since the charge is displaced to a significant length (i.e of the wire) Thus work is done even if we may not be able to easily quantify force.
My questions is this - if the number of parallely-placed wires increases, $W\uparrow$. Thus, we can gain infinite joules by placing more and more parallel wires violating the conservation of energy:
\begin{equation}
\sum_{i=0}^{\infty}W_i = V_i \times1
\end{equation}
by moving the $+1C$ charge in each parallel wire.
How is that possible?
| each wire gets q/n charge (assuming each wire has equal resistance). Moreover Conservation of energy is never violated and why would the parallel combination violate the law. Battery does the work which is equal to qV (q being the total charge accumulated throughout the cycle) and this energy (qv) is used in the whole process no matter what path or system u make
| {
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Is the circuit unitary in Nielsen's method for calculating complexity? I was trying to learn how to calculate circuit complexity (1707.08570) when I chanced upon a seemingly confusing concept. The "Nielsen method" involves looking at a unitary transformation $U$ that takes a reference state $\Psi^R$ to a target state $\Psi^T$:
$$|\Psi^T(t)\rangle=U(t)|\Psi^R(0)\rangle$$
The idea is to rewrite $U$ as an optimized circuit of fundamental quantum gates, and finding the cost associated with this circuit. Examples of these gates include :
\begin{aligned}
H \psi\left(x_{1}, x_{2}\right) &=e^{i \epsilon p_{0} x_{0}} \psi\left(x_{1}, x_{2}\right) & & \text { (global) phase change } \\
J_{1} \psi\left(x_{1}, x_{2}\right) &=\psi\left(x_{1}+\epsilon x_{0}, x_{2}\right) & & \text { shift } x_{1} \text { by constant } \epsilon x_{0} \\
K_{1} \psi\left(x_{1}, x_{2}\right) &=e^{i \epsilon p_{0} x_{1}} \psi\left(x_{1}, x_{2}\right) & & \text { shift } p_{1} \text { by constant } \epsilon p_{0} \\
Q_{21} \psi\left(x_{1}, x_{2}\right) &=\psi\left(x_{1}+\epsilon x_{2}, x_{2}\right) & & \text { shift } x_{1} \text { by } \epsilon x_{2} \quad \text { (entangling gate) } \\
Q_{11} \psi\left(x_{1}, x_{2}\right) &=e^{\epsilon / 2} \psi\left(e^{\epsilon} x_{1}, x_{2}\right) & & \text { scale } x_{1} \rightarrow e^{\epsilon} x_{1} \quad \text { (scaling gate) }
\end{aligned}
where $\epsilon$ is an infinitesimal parameter. In (1810.02734), this procedure leads to the following form of $U$ (Eq 3.38):
$$U(\tau)=\exp \left(\sum_{k=0}^{N-1} \alpha^{k}(\tau) M_{k}^{\text {diag }}\right)$$
where $\{\alpha^k\}$ are complex (and not infinitesimal) and $\{M_k\}$ are diagonal matrices with an identity in the $k^{th}$ position (rest are zero). How can I show that $U$ is indeed unitary as per the original assumption?
Edit : There was a dubious error in the question as was pointed out by @Jahan Claes, regarding unitarity of the gate $Q_{aa}$. I have edited the question accordingly.
| Let's compute $Q_{aa}^\dagger Q_{aa}$. For the rest of this post, I'll drop the $a$ subscript for convenience. Remember that $[x,p]=i$, so that we have $px=xp-i$.
We have:
\begin{align}
Q^\dagger Q & = e^{\epsilon}(e^{i\epsilon xp})^\dagger e^{i\epsilon xp}\\
& = e^{\epsilon}e^{-i\epsilon px}e^{i\epsilon xp}\\
&= e^\epsilon e^{-i\epsilon(xp-i)}e^{i\epsilon xp}\\
&= e^\epsilon e^{-\epsilon}e^{-i\epsilon xp}e^{i\epsilon xp}\\
&=1
\end{align}
So $Q$ is indeed unitary!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/679745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Interpretation of $\oint PdV\neq0$ I hope you are excellent. I'd like you to help me make sense of the integral $ \oint PdV \neq 0 $ for some thermodynamic process. What can it mean for the integral to be nonzero? I can only interpret it as if there is work, however my deep understanding is very limited. I appreciate your comments.
| In thermodynamics, the differential work done on a system is defined as the following:
$$\delta W= -P dV.$$
Work done is an inexact differential, denoted by the symbol $\delta$. This means that the total work depends not only on the initial and endpoints but also on the path taken along the process. To find this total work done for a particular thermodynamic process, one must integrate this differential along the path of the process:
$$W=-\int_{\textrm{process}} P dV.$$
So what does $\oint PdV\neq 0$ mean? This is simply a statement of the path-dependent property of work done. You can see this by breaking up the loop integral into two integrals.
$$\oint PdV=\int_1^2PdV+\int_2^1PdV ,$$
where each integral is taken along a different path.
If work done did not depend on the path taken, then we could simply write the following, as a property of integrals:
$$\int_1^2PdV=-\int_2^1PdV \tag{1}$$
yielding,
$$\oint PdV=0.$$
However, since work done depends on the path taken, equation $(1)$ is not true in general, thus
$$\oint PdV \neq 0.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/679868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the force on a block if another block sits on it? So imagine there is a Block A on top of a block B on Earth's crust. What is the force acting on Block B from Block A? Shouldn't it be the force of gravity acting on Block A because that force is pushing down on Block B?
| Gravity is an attractive force, so we tend to think of it as "pulling" rather than "pushing". But you're right that block B experiences a gravitational force from A.
There is a second force. Block A doesn't fall through Block B, even though there's an attractive gravitational force – according to Newton's Second Law, it seems like it should. There must be a force acting on block A from block B (a contact force) to stop it accelerating. We call this the normal reaction force. You can think of it as the force that atoms experience not wanting to be squashed together. By Newton's Third Law, that means there's also a force from block A on block B (confusingly, the "equal and opposite reaction" to the "normal reaction").
If the blocks were electrically charged, there would be a further force, separate to these two.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/680037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why do we take account of the whole solid sphere when calculating potential energy of a point inside a solid sphere? See I know that Newton's shell theorem says that any point inside a spherical shell does not encounter any gravitational force by the (outer)shell and it is zero. The same principle we use while finding the weight of a body deep inside Earth. But when calculating the potential of a point inside a solid sphere we always account for the potential due to the outer shells as well. Why is that?
| The shell theorem relies on the fact that force is a vector, and so the vector sum of forces cancels out inside each shell.
Potential energy is a scalar, and more importantly it is the same sign for all contributions from a given shell. Therefore, the potential energy does not cancel out for each shell and must be considered.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/680156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If I lift a body with a force greater than its weight, what will happen to the excess energy provided to the body I will give an example to explain my question.
Case 1:
An elevator lifts body a with force equal to its weight for a distance $d$
*
*Energy given to the body (work done)$=$ Weight $×$ $d$
*Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ Weight $× \ d$
Case 2:
An elevator lifts the same body with force equal to twice weight it’s for a distance $d$
*
*Energy given to the body (work done) $=$ $2 \ ×$ weight $×$ $d$
*Amount of work the body is capable of doing by falling down (gravitational potential energy) $=$ weight $×$ $d$
So doubled the amount of energy I gave to the body yet it’s capacity to do work by falling down has not changed.
Where is the excess energy the lift provided the body?
(I am in 11th grade so please make your explanation simple enough for me to understand.)
| In both cases, the work provided by the elevator is turned into kinetic energy of the object.
In the moment the elevator stops (and stops doing work), the object thus carries kinetic energy and therefor will continue flying upwards. Both of your potential-energy calculations are therefore wrong - the actual top height will be more than $d$.
And the object will naturally reach higher in the second case since it gains a larger amount of kinetic energy in the second case, causing more potential energy to be stored.
In regular elevators you might feel various lifting forces without "flying" upwards as the elevator stops. So you might feel that the answer I provided is incorrect. But remember to include the decceleration as well. The larger lifting force in certain elevators might be exerted for a shorter time as well, with deceleration beginning before the stop is reached. In the ideal scenario of the elevator keeping a constant speed and then suddenly stopping immediately, then you will experience the "flying".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/680295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there any *global* timelike Killing vector in Schwarzschild geometry? I have been dealing with the following issue related to the Schwarzschild geometry recently. When expressed as:
$$
ds^{2}=-\left(1-\frac{2GM}{r}\right)dt^{2}+\frac{1}{1-\frac{2GM}{r}}dr^{2}+d\Omega_{2}^{2}$$
one can find a Killing vector $\xi=\partial_{t}$, since there are no components of the metric depending on $t$. This Killing vector is timelike for $r>2GM$, but spacelike for $r<2GM$ (since $\xi^{\mu}\xi_{\mu}=-\left(1-\frac{2GM}{r}\right)$). My question is:
*
*Can we find any timelike vector for the region $r<2GM$?
*If not, this would imply that the Schwarzschild solution is not stationary for $r<2GM$. But it is usually referred to as a "static spacetime". This wouldn't be true for the region $r<2GM$. So is this an abuse of language?
| A time-line coordinate flips only if one assumes that Schwarzschild's vacuum solution is valid in the interior of a black hole, too. In my view, it is not correct. A solution of Einstein field equations spans the whole spacetime. In case of a constant energy density in some region of spacetime (step function) the solution (metric) can be cast into two parts: an interior and an exterior one. It is admissible to work with only one of them, and glue it to each other, but it is not admissible to extend them over their validity domain. For example the Schwarzschild interior solution remains static, although maybe not stable, even above the critical compactness parameter $\alpha=8/9$. This view gave rise to Pawel O. Mazur and Emil Mottola idea of gravastar, https://en.wikipedia.org/wiki/Gravastar . Their paper you can read here: https://arxiv.org/abs/1501.03806.
If you like see also: https://physics.stackexchange.com/a/679431/281096 and https://physics.stackexchange.com/a/674311/281096 .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/680739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
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